Question

| 27. If sin (A + B) = sin A cos B+cos Asin B
and cos (A - B) = cos A cos B+sin A sin B,
find the values of (i) sin 75° and (ii) cos 15°​

Answer 1

Given identities are,

$\large\sin(A+B)=\sin A\cos B+cos A\sin B\\ \\ \\ \large\cos(A-B)=\cos A\cos B+\sin A\sin B$

Using these, we have to find,

(i)  sin 75°

(ii)  cos 15°

But before, we have to recall some trigonometric values which are given below.

→   sin 30° = cos 60° = 1/2

→   sin 45° = cos 45° = 1/√2

→   sin 60° = cos 30° = √3/2

So we may calculate using these values.

And one more, we must recall this.

sin α = cos (90 - α)

So, add the given values in the question. We get 90°.

75° + 15° = 90°

So we can say,

sin 75° = cos 15°

(i)   sin 75°

⇒  sin (30° + 45°)

⇒  sin 30° cos 45° + cos 30° sin 45°

⇒  (1/2) · (1/√2) + (√3/2) · (1/√2)

⇒  (1/√2) · ((1/2) + (√3/2))

⇒  (1/√2)((1 + √3)/2)

⇒  (1 + √3) / 2√2

⇒  cos 15°

Hence,

sin 75° = cos 15° = (1 + √3) / 2√2

Now we can substitute the approximate values of the irrationals.

→   √2 ≈ 1.414

→   √3 ≈ 1.732

So,

    (1 + √3) / 2√2

⇒  (1 + 1.732) / (2 · 1.414)

⇒  2.732 / 2.828

⇒  0.9661

Answer 2

Trigonometric Identities :

sin ( A + B) = sinA cos B + cosA sinB

cos(A - B ) = cosA cosB + sinA sinB

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1 ). sin 75°

Split sin ( 75° ) as sin ( 45° + 30°)

Let A = 45°, B = 30°

Using formula of sin ( A + B ) ,

sin ( 45° + 30°) = sin 45°. cos30° + cos 45°. sin 30°

sin ( 45° + 30°) = $\mathsf{\dfrac{1}{\sqrt{2}}\:*\:{\dfrac{\sqrt{3}}{2}\:+\:{\dfrac{1}{\sqrt{2}}\:*\:{\dfrac{1}{2}}}}}$

⛛ sin ( 45° + 30°) = $\mathsf{\dfrac{\sqrt{3}\:+\:1}{</strong><strong>2</strong><strong>{</strong><strong>\</strong><strong>sqrt</strong><strong>{</strong><strong>2</strong><strong>}}</strong><strong>}</strong><strong>}</strong><strong>$

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2 ). cos 15°

Using above formula of cos ( A - B),

Let A = 45° and B = 15°

cos ( 45° - 30°) = cos 45°. cos30° + sin 45°. sin 30°

cos ( 45° - 30° ) = $\mathsf{\dfrac{1}{\sqrt{2}}\:*\:{\dfrac{\sqrt{3}}{2}\:+\:{\dfrac{1}{\sqrt{2}}\:*\:{\dfrac{1}{2}}}}}$

⛛ cos ( 45° - 30° ) = $\mathsf{\dfrac{\sqrt{3}\:+\:1}{2{\sqrt{2}}}}$

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Since, sin( 90 - 15)° = cos 15° .

⛛ So, the answer of both parts is same.