Question

Find the value of sin 75°

Answer 1

Hint: To solve this question, we will make use of trigonometric ratios of compound angles. First of all, we will understand what are the compound angles and what are the various trigonometric ratios of compound angles. Then we will select the trigonometric ratio sine of compound angles. Thus, we can find the value of sin 75°.

Complete step by step answer:
An angle is called a compound angle when that angle is made by addition or subtraction of two angles.
For example, if one of the angle is ∠A and the other angle is ∠B, then ∠(A + B) is an example of compound angle. Another compound angle that can be made from ∠A and ∠B can be ∠(A – B).
Trigonometric ratios of compound angles do not have associate property.
For example, tan (A + B) $\ne $ tan A + tan B.
There are special formulas for compound angles. Formula for compound angles of sine, cosine and tangent trigonometric ratios are given as follows:
sin (A + B) = sin A $\times $ cos B + cos A $\times $ sin B
sin (A ─ B) = sin A $\times $ cos B + cos A $\times $ sin B
cos (A + B) = cos A $\times $ cos B ─ sin A $\times $ sin B
sin (A ─ B) = cos A $\times $ cos B + sin A $\times $ sin B
tan (A + B) = $\dfrac{\operatorname{tanA}+tanB}{1-\operatorname{tanAtanB}}$
tan (A ─ B) = $\dfrac{\operatorname{tanA}-tanB}{1+\operatorname{tanAtanB}}$
We are supposed to find the value of sin 75°. 75 can also be written as 30 + 45.
$\Rightarrow $ sin 75° = sin (30° + 45°)
We will apply trigonometric ratios of compound angles.
$\Rightarrow $ sin 75° = sin 30° $\times $ cos 45° + cos 30° $\times $ sin 45°
We know that sin 30° = $\dfrac{1}{2}$, cos 45° = sin 45° = $\dfrac{1}{\sqrt{2}}$ and cos 30° = $\dfrac{\sqrt{3}}{2}$.
$\Rightarrow $ sin 75° = $\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}$

$\Rightarrow $ sin 75° = $\dfrac{1+\sqrt{3}}{2\sqrt{2}}$

Note: Students are advised to by heart the formulas for trigonometric ratios of compound angles. Students should know conversion formulas as well for these types of problems.