Q) It is given that sin (A – B) = sin A cos B – cos A sin B. Use it to find the value of sin 15°.

PYQ: Q 23(a) – CBSE 2025 – Code 30 – Series 5 – Set 1

Ans: 

Step 1: Here, we need to find value of sin 15°.

Let’s split 15° into two angles in such a way that we can put values of their sin and cos forms.

Therefore, we write 15° as: 15° = 45° − 30°

Step 2:
Since, we are given that

sin (A – B) = sin A cos B – cos A sin B

∴ sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°

Step 3: Let’s put the values of sine and cosine forms for 45° and 30°:

We have, sin 45° = 1/√2, cos 45° = 1/√2, sin 30° = 1 / 2 and cos 30° = √3 / 2

Let’s substitute these values into the given identity:

∴ sin 15° = (1 / √2) (√3 / 2) − (1 / √2) (1 / 2)

∴ sin 15° = √3 / 2√2 − 1/ 2 √2

∴ sin 15° = (√3 − 1) / 2 √2

If we multiply the above equation, by √2 in its numerator & denominator, we get:

∴ sin 15° = (√3 − 1) (√2) / (2 √2) (√2)

∴ sin 15° = (√6 − √2) / 4

Therefore, the value of sin 15° is (√6 − √2) / 4

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