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Q.

If sinA+sin2A+sin3A=1, then find the value of cos6A-4cos4A+8cos2A.


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a

1

b

2

c

3

d

4  

answer is D.

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Detailed Solution

It is given, sinA+sin2A+sin3A=1.
sinA+sin2A+sin3A=1 ⇒sinA(1+sin2A)=1-sin2A ⇒sinA(2-cos2A)=cos2A Squaring on both sides, we get
⇒sin2A(4-4cos2A+cos4A)=cos4A  ⇒1-cos2A4-4cos2A+cos4A=cos4A ⇒4-8cos2A+5cos4A-cos6A=cos4A ∴cos6A-4cos4A+8cos2A=4.
Hence, the correct option is (4).
 
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