Submultiple Angles

Submultiple angles in trigonometry refer to angles that are a fraction of a given angle, typically a half or a third. They are used in trigonometric identities and equations where a larger angle is expressed as a multiple of a smaller one. For example, if \(\theta\) is an angle, then \(\theta/2\) and \(\theta/3\) are its submultiples. Trigonometric functions involving submultiple angles (like \(\sin(\theta/2)\), \(\cos(\theta/3)\), etc.) are important in solving various trigonometric problems and are often used in deriving certain identities and equations.

1)

\[ \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]
Proof:

We will prove this using the double angle formula for sine, which states:

\[ \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \]

Let \(\alpha = \frac{\theta}{2}\). Then the double angle formula becomes:

\[ \sin(2 \times \frac{\theta}{2}) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]

Simplifying the left-hand side, we have:

\[ \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]

\(\blacksquare\)

2)

\[ \cos(\theta) = \cos^2\left(\frac{\theta}{2}\right) - \sin^2\left(\frac{\theta}{2}\right) = 2\cos^2\left(\frac{\theta}{2}\right) - 1 = 1 - 2\sin^2\left(\frac{\theta}{2}\right) \]
Proof:

We will use the cosine double angle formula, \(\cos(2\alpha) = 2\cos^2(\alpha) - 1\) and \(\cos(2\alpha) = 1 - 2\sin^2(\alpha)\), replacing \(\alpha\) with \(\frac{\theta}{2}\) to prove this theorem.

  1. Using Cosine Double Angle Formula:

    \[ \cos(\theta) = \cos(2 \times \frac{\theta}{2}) = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \]

  2. Alternatively, Using the Other Form of Cosine Double Angle Formula:

    \[ \cos(\theta) = \cos(2 \times \frac{\theta}{2}) = 1 - 2\sin^2\left(\frac{\theta}{2}\right) \]

  3. Using the Identity \(\cos^2(\alpha) - \sin^2(\alpha) = \cos(2\alpha)\):

    \[ \cos(\theta) = \cos^2\left(\frac{\theta}{2}\right) - \sin^2\left(\frac{\theta}{2}\right) \]

3)

\[ \tan(\theta) = \frac{2\tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} \]
Proof:

The formula for the tangent of a sum, \(\tan(\alpha + \beta)\), is:

\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \]

Let \(\alpha = \beta = \frac{\theta}{2}\). Then, we get:

\[ \tan(\theta) = \tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\theta}{2}\right) + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)\tan\left(\frac{\theta}{2}\right)} \]
\[ = \frac{2\tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} \]

\(\blacksquare\)

4)

\[ \cot(\theta) = \frac{\cot^2\left(\frac{\theta}{2}\right) - 1}{2\cot\left(\frac{\theta}{2}\right)} \]
Proof:

The formula for the cotangent of a sum \(\alpha + \beta\) is:

\[ \cot(\alpha + \beta) = \frac{\cot(\alpha)\cot(\beta) - 1}{\cot(\beta) + \cot(\alpha)} \]

Let \(\alpha = \beta = \frac{\theta}{2}\). Then, we have:

\[ \cot(\theta) = \cot\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{\cot^2\left(\frac{\theta}{2}\right) - 1}{\cot\left(\frac{\theta}{2}\right) + \cot\left(\frac{\theta}{2}\right)} \]
\[ = \frac{\cot^2\left(\frac{\theta}{2}\right) - 1}{2\cot\left(\frac{\theta}{2}\right)} \]

\(\blacksquare\)

5) Expressing trigonometrical ratios of the angle \(\frac{\theta}{2}\) in terms of \(\cos\theta\).

The expressions for \(\cos\left(\frac{\theta}{2}\right)\), \(\sin\left(\frac{\theta}{2}\right)\), and \(\tan\left(\frac{\theta}{2}\right)\) in terms of \(\cos(\theta)\) are given by:

  1. \(\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos(\theta)}{2}}\)
  2. \(\sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos(\theta)}{2}}\)
  3. \(\tan\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}}\)

The plus or minus sign is determined based on the quadrant of \(\frac{\theta}{2}\).

Proof:

  1. For \(\cos\left(\frac{\theta}{2}\right)\): From the identity

    \[ \cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \]

    we rearrange to find \(\cos\left(\frac{\theta}{2}\right)\):

    \[ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos(\theta)}{2} \]

    Taking the square root (and considering the quadrant for the correct sign):

    \[ \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos(\theta)}{2}} \]

  2. For \(\sin\left(\frac{\theta}{2}\right)\): Using the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\), we have

    \[ \sin^2\left(\frac{\theta}{2}\right) = 1 - \cos^2\left(\frac{\theta}{2}\right) \]

    Substitute \(\cos^2\left(\frac{\theta}{2}\right)\) from the previous step:

    \[ \sin^2\left(\frac{\theta}{2}\right) = 1 - \frac{1 + \cos(\theta)}{2} = \frac{1 - \cos(\theta)}{2} \]

    Taking the square root (considering the quadrant for the correct sign):

    \[ \sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos(\theta)}{2}} \]

  3. For \(\tan\left(\frac{\theta}{2}\right)\): Using the definition of tangent in terms of sine and cosine:

    \[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \]

    Substitute the expressions for \(\sin\left(\frac{\theta}{2}\right)\) and \(\cos\left(\frac{\theta}{2}\right)\):

    \[ \tan\left(\frac{\theta}{2}\right) = \frac{\pm\sqrt{\frac{1 - \cos(\theta)}{2}}}{\pm\sqrt{\frac{1 + \cos(\theta)}{2}}} \]

    Simplifying, we get:

    \[ \tan\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \]

This completes the proof of the theorem. The correct sign in each case depends on the quadrant in which \(\frac{\theta}{2}\) lies.

Example

Suppose \(\theta=\frac{17\pi}{3}\). Find \(\cos\theta\). Then find \(\cos\frac{\theta}{2}\), \(\sin\frac{\theta}{2}\) and \(\tan\frac{\theta}{2}\).

Solution:

  1. Cosine of \(\theta\):

    \[ \theta = \frac{17\pi}{3} = 6\pi - \frac{\pi}{3} \]

    In this form, \(\frac{17\pi}{3}\) is in the 4th quadrant, where the cosine is positive. Therefore:

    \[ \cos(\theta) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

  2. Cosine of \(\frac{\theta}{2}\):

    \[ \frac{\theta}{2} = \frac{17\pi}{6} = 3\pi - \frac{\pi}{6} \]

    This places \(\frac{\theta}{2}\) in the 2nd quadrant, where the cosine is negative. Using the formula:

    \[ \cos\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 + \cos(\theta)}{2}} = -\sqrt{\frac{1 + \frac{1}{2}}{2}} = -\frac{\sqrt{3}}{2} \]

  3. Sine of \(\frac{\theta}{2}\): Since \(\frac{\theta}{2}\) is in the 2nd quadrant, where the sine is positive:

    \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} = \sqrt{\frac{1 - \frac{1}{2}}{2}} = \frac{1}{2} \]

  4. Tangent of \(\frac{\theta}{2}\): Given \(\frac{\theta}{2}\) is in the 2nd quadrant, where the tangent is negative:

    \[ \tan\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} = -\sqrt{\frac{1 - \frac{1}{2}}{1 + \frac{1}{2}}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \]

This completes the evaluation using the correct quadrant for \(\frac{\theta}{2}\).

Example

Now suppose \(\theta=\frac{17\pi}{6}\). Find \(\cos\theta\). Then find \(\cos\frac{\theta}{2}\), \(\sin\frac{\theta}{2}\) and \(\tan\frac{\theta}{2}\).

Solution

  1. Determining the Quadrant of \(\theta\):

    [ \theta = \frac{17\pi}{6} = 3\pi + \frac{\pi}{6} ] This places \(\theta\) in the 3rd quadrant. In the 3rd quadrant, both cosine and sine are negative.

  2. Cosine of \(\theta\):

    Since \(\frac{17\pi}{6}\) is equivalent to \(3\pi + \frac{\pi}{6}\), the cosine value will be the same as the cosine of \(\frac{\pi}{6}\), but negative due to the quadrant:

    \[ \cos(\theta) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

  3. Determining the Quadrant of \(\frac{\theta}{2}\):

    \[ \frac{\theta}{2} = \frac{17\pi}{12} = \pi + \frac{5\pi}{12} \]

    This places \(\frac{\theta}{2}\) in the 3rd quadrant as well. In the 3rd quadrant, both cosine and sine are negative.

  4. Cosine of \(\frac{\theta}{2}\):

    Since \(\frac{\theta}{2}\) is in the 3rd quadrant, where cosine is negative:

    \[ \cos\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 + \cos(\theta)}{2}} = -\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \]
    \[ = -\sqrt{\frac{2 - \sqrt{3}}{4}} \]
    \[ = -\sqrt{\frac{4 - 2\sqrt{3}}{8}} \]
    \[ = -\sqrt{\frac{(\sqrt{3} - 1)^2}{8}} \]
    \[ = -\frac{\sqrt{3} - 1}{2\sqrt{2}} \]

  5. Sine of \(\frac{\theta}{2}\):

    As \(\frac{\theta}{2}\) is in the 3rd quadrant, where sine is negative:

    \[ \sin\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 - \cos(\theta)}{2}} = -\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \]
    \[ = -\sqrt{\frac{2 + \sqrt{3}}{4}} \]
    \[ = -\sqrt{\frac{4 + 2\sqrt{3}}{8}} \]
    \[ = -\sqrt{\frac{(\sqrt{3} + 1)^2}{8}} \]
    \[ = -\frac{\sqrt{3} + 1}{2\sqrt{2}} \]

  6. Tangent of \(\frac{\theta}{2}\):

    Given \(\frac{\theta}{2}\) is in the 3rd quadrant, where tangent is positive:

    \[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} \]
    \[ = \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}} \]
    \[ = \sqrt{\frac{2(2 + \sqrt{3})}{2(2 - \sqrt{3})}} \]
    \[ = \sqrt{\frac{4 + 2\sqrt{3}}{4 - 2\sqrt{3}}} \]
    \[ = \sqrt{\frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)^2}} \]
    \[ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]
    \[ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \]
    \[ = \frac{3 + 2\sqrt{3} + 1}{3 - 1} \]
    \[ = \frac{4 + 2\sqrt{3}}{2} \]
    \[ = 2 + \sqrt{3} \]

6)

\[ (\sin(\theta/2) \pm \cos(\theta/2))^2 = 1 \pm \sin(\theta) \]
Proof:

Expanding the left-hand side and using trigonometric identities, we have:

\[ \sin^2(\theta/2) \pm 2\sin(\theta/2)\cos(\theta/2) + \cos^2(\theta/2) \]
\[ = 1 \pm 2\sin(\theta/2)\cos(\theta/2) \]

(Using \(\sin^2(\alpha) + \cos^2(\alpha) = 1\))

Applying the double angle formula for sine (\(\sin(\theta) = 2\sin(\theta/2)\cos(\theta/2)\)):

\[ = 1 \pm \sin(\theta) \]

This completes the proof.

7)

\[ \sin\left(\frac{\theta}{2}\right) = \frac{\left(\pm\sqrt{1 + \sin(\theta)}\right) + \left(\pm\sqrt{1 - \sin(\theta)}\right)}{2} \]
\[ \cos\left(\frac{\theta}{2}\right) = \frac{\left(\pm\sqrt{1 + \sin(\theta)}\right) - \left(\pm\sqrt{1 - \sin(\theta)}\right)}{2} \]
Proof:
\[ \sin\left(\frac{\theta}{2}\right) + \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{1 + \sin(\theta)} \tag{1} \]
\[ \sin\left(\frac{\theta}{2}\right) - \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{1 - \sin(\theta)} \tag{2} \]

By adding equations (1) and (2), we find \( 2\sin(\theta/2) \):

\[ 2\sin\left(\frac{\theta}{2}\right) = \left(\pm\sqrt{1 + \sin(\theta)}\right) + \left(\pm\sqrt{1 - \sin(\theta)}\right) \]
\[ \sin\left(\frac{\theta}{2}\right) = \frac{\left(\pm\sqrt{1 + \sin(\theta)}\right) + \left(\pm\sqrt{1 - \sin(\theta)}\right)}{2} \]

By subtracting equation (2) from (1), we find \( 2\cos(\theta/2) \):

\[ 2\cos\left(\frac{\theta}{2}\right) = \left(\pm\sqrt{1 + \sin(\theta)}\right) - \left(\pm\sqrt{1 - \sin(\theta)}\right) \]
\[ \cos\left(\frac{\theta}{2}\right) = \frac{\left(\pm\sqrt{1 + \sin(\theta)}\right) - \left(\pm\sqrt{1 - \sin(\theta)}\right)}{2} \]

These results give us \( \sin(\theta/2) \) and \( \cos(\theta/2) \) in terms of \( \sin(\theta) \).

How to choose the signs in the above formula

Above we explained the formula which expresses \(\sin\frac{\theta}{2}\) and \(\cos\frac{\theta}{2}\) in terms of \(sin\theta\), but we do not still know about how to decide those signs.

\[ \sin\left(\frac{\theta}{2}\right) + \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{1 + \sin(\theta)} \tag{1} \]
\[ \sin\left(\frac{\theta}{2}\right) - \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{1 - \sin(\theta)} \tag{2} \]

Equations (1) and (2) in the proof are the key equations to decide the correct signs. in eqn (1) we decide whether the the expression on the left \(\sin\left(\frac{\theta}{2}\right) + \cos\left(\frac{\theta}{2}\right)\) is positive or negative depending on the value or some information about \(\frac{\theta}{2}\).

sign of sin theta by 2 plus cos theta by 2

Draw a unit circle as in the above figure. Draw a line \(PQ\) making an angle of \(\frac{\pi}{4}\) with negative x-axis. Thsi line divides the circle into two regions. If the terminal point of angle \(\frac{\theta}{2}\) lies the blue region, then \(\sin\left(\frac{\theta}{2}\right) + \cos\left(\frac{\theta}{2}\right)\) is positive and if lies in the red region, then it is negative.

The explanation of this fact is quite simple and we leave it as an exercise for you to verify.

Similarly we decide the sign of \(\sin\left(\frac{\theta}{2}\right) - \cos\left(\frac{\theta}{2}\right)\) of the left hand side of eqn(2) using the figure below.

sign of sin theta by 2 minus sign theta by 2

8) The trigonometric functions \( \sin(\theta) \), \( \cos(\theta) \), and \( \tan(\theta) \) can be expressed in terms of \( \tan(\theta/2) \) using the tangent half-angle formulas:

  • Sine in terms of \( \tan(\theta/2) \):

    \[ \sin(\theta) = \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} \]

  • Cosine in terms of \( \tan(\theta/2) \):

    \[ \cos(\theta) = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} \]

  • Tangent in terms of \( \tan(\theta/2) \):

    \[ \tan(\theta) = \frac{2\tan(\theta/2)}{1 - \tan^2(\theta/2)} \]

These formulas come from the double angle and half-angle formulas in trigonometry, where \( \tan(\theta/2) \) is often denoted as \( t \) when simplifying expressions or solving trigonometric equations.

9) The trigonometric functions \( \sin(\theta) \), \( \cos(\theta) \), and \( \tan(\theta) \) can be expressed in terms of \( \sin\left(\frac{\theta}{3}\right) \), \( \cos\left(\frac{\theta}{3}\right) \), and \( \tan\left(\frac{\theta}{3}\right) \) respectively, as follows:

  • \( \sin(\theta) \) can be expressed as:

    \[ \sin(\theta) = 3\sin\left(\frac{\theta}{3}\right) - 4\sin^3\left(\frac{\theta}{3}\right) \]

  • \( \cos(\theta) \) can be expressed as:

    \[ \cos(\theta) = 4\cos^3\left(\frac{\theta}{3}\right) - 3\cos\left(\frac{\theta}{3}\right) \]

  • \( \tan(\theta) \) can be expressed as:

    \[ \tan(\theta) = \frac{3\tan\left(\frac{\theta}{3}\right) - \tan^3\left(\frac{\theta}{3}\right)}{1 - 3\tan^2\left(\frac{\theta}{3}\right)} \]
Proof:

Let \( \alpha = \frac{\theta}{3} \). We will use the triple-angle formulas to prove the identities for \( \sin(\theta) \), \( \cos(\theta) \), and \( \tan(\theta) \).

  1. For \( \sin(\theta) \): The triple-angle formula for sine is given by:
\[ \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha) \]

Substituting \( \alpha = \frac{\theta}{3} \) gives us:

\[ \sin(\theta) = 3\sin\left(\frac{\theta}{3}\right) - 4\sin^3\left(\frac{\theta}{3}\right) \]
  1. For \( \cos(\theta) \): The triple-angle formula for cosine is:
\[ \cos(3\alpha) = 4\cos^3(\alpha) - 3\cos(\alpha) \]

Substituting \( \alpha = \frac{\theta}{3} \) gives us:

\[ \cos(\theta) = 4\cos^3\left(\frac{\theta}{3}\right) - 3\cos\left(\frac{\theta}{3}\right) \]
  1. For \( \tan(\theta) \): The triple-angle formula for tangent, derived from the sine and cosine triple-angle formulas, is:
\[ \tan(3\alpha) = \frac{3\tan(\alpha) - \tan^3(\alpha)}{1 - 3\tan^2(\alpha)} \]

Substituting \( \alpha = \frac{\theta}{3} \) gives us:

\[ \tan(\theta) = \frac{3\tan\left(\frac{\theta}{3}\right) - \tan^3\left(\frac{\theta}{3}\right)}{1 - 3\tan^2\left(\frac{\theta}{3}\right)} \]

Example

Find the product \( \cos(10^\circ)\cos(130^\circ)\cos(250^\circ) \).

Solution:

To prove we use the triple angle formula:

\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]

If we set \( \theta = 10^\circ \), \( 3\theta = 30^\circ \), and we know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so the equation becomes:

\[ \frac{\sqrt{3}}{2} = 4\cos^3(10^\circ) - 3\cos(10^\circ) \]
\[ 0 = 4\cos^3(10^\circ) - 3\cos(10^\circ) - \frac{\sqrt{3}}{2} \]

If we set \( \theta = 130^\circ \), \( 3\theta = 390^\circ \), which is equivalent to \( 30^\circ \) because cosine is a periodic function with a period of \( 360^\circ \). Thus, we have:

\[ \frac{\sqrt{3}}{2} = 4\cos^3(130^\circ) - 3\cos(130^\circ) \]
\[ 0 = 4\cos^3(130^\circ) - 3\cos(130^\circ) - \frac{\sqrt{3}}{2} \]

Similarly, if \( \theta = 250^\circ \), \( 3\theta = 750^\circ \), which again reduces to \( 30^\circ \) after subtracting full cycles of \( 360^\circ \), we get:

\[ \frac{\sqrt{3}}{2} = 4\cos^3(250^\circ) - 3\cos(250^\circ) \]
\[ 0 = 4\cos^3(250^\circ) - 3\cos(250^\circ) - \frac{\sqrt{3}}{2} \]

Each of these is a cubic equation in \( \cos(\theta) \) and they all have the same structure. Since \( \cos(10^\circ) \), \( \cos(130^\circ) \), and \( \cos(250^\circ) \) are distinct, they are the roots of the cubic equation:

\[ 4x^3 - 3x - \frac{\sqrt{3}}{2} = 0 \]

Using Vieta's formulas, the product of the roots of a cubic equation is \( -\frac{d}{a} \). Hence, the product of the roots \( \cos(10^\circ) \), \( \cos(130^\circ) \), and \( \cos(250^\circ) \) is:

\[ \cos(10^\circ) \cdot \cos(130^\circ) \cdot \cos(250^\circ) = -\left(-\frac{\sqrt{3}}{2}\right) \div 4 \]
\[ \cos(10^\circ) \cdot \cos(130^\circ) \cdot \cos(250^\circ) = \frac{\sqrt{3}}{8} \]

Therefore, the product \( \cos(10^\circ)\cos(130^\circ)\cos(250^\circ) \) equals \( \frac{\sqrt{3}}{8} \).