If sinx+sin2x=1, then the value of expression cos12x+3cos10x+3cos8x+cos6x-1 is equal to
0
1
-1
2
Explanation for the correct option
Step 1: Given information
∵sinx+sin2x=1
⇒sinx=1-sin2x
⇒sinx=cos2xOrcos2x=sinx.......(1)
Step 2: According to the question solve the given expression
∵cos12x+3cos10x+3cos8x+cos6x–1
⇒cos4x3+3cos4x2cos2x+3cos4xcos2x2+cos2x3–1
⇒sin2x3+3sin2x2cos2x+3sin2xcos2x2+cos2x3–1 (From equation (1))
Using the formula ∵(a+b)3=a3+3a2b+3ab2+b3
∴(sin2x+cos2x)3-1
⇒(1)3-1(sin2x+cos2x=1)
⇒1-1⇒0
Hence option (A) is the correct answer.
The maximum value of f(x)=sin2x1+cos2xcos2x1+sin2xcos2xcos2xsin2xcos2xsin2x,x∈R is: