Question

Suppose sin⁡2θ=cos⁡3θ, here 0<θ<π2 then what is the value of cos⁡2θ?

• A. (A) 1+54
• B. (B) 1−258
• C. (C) −1+54
• D. (D) −1−54

Answer 1

Answer: (A)

(A) 1+54

Answer 2

Explanation:
Given:sin⁡2θ=cos⁡3θ,0<θ<π2As we know that, sin⁡2θ=2sin⁡θcos⁡θ and cos⁡3θ=4cos3⁡θ−3cos⁡θ⇒2sin⁡θcos⁡θ=4cos3⁡θ−3cos⁡θ⇒2sin⁡θ=4cos2⁡θ−3⇒2sin⁡θ=4(1−sin2⁡θ)−3=4−4sin2⁡θ−3⇒4sin2⁡θ+2sin⁡θ−1=0Comparing the above equation with quadratic equation ax2+bx+c=0,a=4,b=2 and c=−1Now substituting the values in the quadratic formula x=(−b±b2−4ac)2a we get,sin⁡θ=−2±−22−4(4)(−1)2(4)=−2±4+168=−2±208=−1±54Thus, sin⁡θ=−1±54Since, 0<θ<π2⇒θ lies between 0∘ to 90∘⇒ all ratios are positive.⇒sin⁡θ=−1+54As we know that, cos⁡2θ=cos2⁡θ−sin2⁡θ=1−2sin2⁡θ⇒cos⁡2θ=1−2(−1+54)2=1+54∴ The the value of cos⁡2θ is 1+54Hence, the correct option is (A).