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If sin alpha = 1/2 than prove that ( 3 cos alpha - 4 cos^3 alpha ) = 0?
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If sin alpha = 1/2 than prove that ( 3 cos alpha - 4 cos^3 alpha ) = 0...
Proof that (3cosα - 4cos³α) = 0 when sinα = 1/2

Given, sinα = 1/2
We need to prove that (3cosα - 4cos³α) = 0

Using Trigonometric Identity

We know that sin²α + cos²α = 1 (Trigonometric Identity)
Substituting sinα = 1/2 in this equation, we get
(1/2)² + cos²α = 1
Simplifying it, we get
cos²α = 3/4
Taking square root on both sides, we get
cosα = ±√3/2

Checking for cosα = √3/2

If cosα = √3/2, then
3cosα - 4cos³α = 3(√3/2) - 4(√3/2)³
= 3(√3/2) - 4(3√3/8)
= (3√3/2) - (3√3/2)
= 0
Hence, for cosα = √3/2, (3cosα - 4cos³α) = 0

Checking for cosα = -√3/2

If cosα = -√3/2, then
3cosα - 4cos³α = 3(-√3/2) - 4(-√3/2)³
= -3(√3/2) - 4(-27√3/8)
= -3(√3/2) + 27√3/2
= 12√3/2
This is not equal to 0.
Hence, for cosα = -√3/2, (3cosα - 4cos³α) ≠ 0

Conclusion

Thus, we have proved that (3cosα - 4cos³α) = 0 when sinα = 1/2 only for cosα = √3/2.
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If sin alpha = 1/2 than prove that ( 3 cos alpha - 4 cos^3 alpha ) = 0?
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