Question

If sin a=1/2,show that (3vos a-4cos cube a) =0
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Answer 1

Question :

• If sin A = 1/2 , show that ( 3 cos A - 4cos³ A ) =0

Solution :

sin A = 1 / 2

So, cos A = √3 / 2

cos³ A = ( √3 / 2 )³

cos³ A = 3√3 / 8

Now

3cos A - 4cos³ A

→ 3 × √3 / 2 – 4 × 3√3 / 8

→ 3√3 / 2 ) – ( 3√3 / 2 )

→ 0

LH.S = R.H.S

Hence proved

Answer 2

Given that ,

Sin(a) = 1/2 = P/H

By pythagoras theorem ,

(H)² = (B)² + (P)²

(2)² = (B)² + (1)²

4 = (B)² + 1

(B)² = 3

B = √3 unit

Now , we have to prove 3Cos(a) - 4Cos³(a) = 0

Thus ,

$\sf \mapsto 3( \frac{ \sqrt{3} }{2} ) - 4 { (\frac{ \sqrt{3} }{2})}^{3} \\ \\ \sf \mapsto \frac{3 \sqrt{3} }{2} - \frac{12\sqrt{3} }{8} \\ \\ \sf \mapsto \frac{3 \sqrt{3} - 3 \sqrt{3} }{2} \\ \\ \sf \mapsto0$

Hence proved