Question

The sin (theta) = -2/5, and theta lies in quadrant IV. Find the exact values of the sine and cosine of 2 theta.

Answer 1

`\displaystyle\\Answer:\ sin(2\theta)=-(4√(21) )/(25) ,\ cos(2\theta)=(17)/(25)`

Explanation:

`\displaystyle\\sin(\theta)=-(2)/(5) \ \ \ \ \ \ \ \ 270^0 < \theta < 360^0\\\\sin^2(\theta)+cos^2(\theta)=1\\\\cos^2(\theta)=1-sin^2(\theta)\\\\Hence,\\\\cos^2(\theta)=1-(-(2)/(5))^2 \\cos^2(\theta)=1-(4)/(25) \\\\cos^2(\theta)=(25(1)-4)/(25) \\\\cos^2(\theta)=(21)/(25) \\\\`

Extract the square root of both parts of the equation:

`\displaystyle\\cos(\theta)=б\sqrt{(21)/(25) } \\\\cos(\theta)=б(√(21) )/(5) \\\\270^0 < \theta < 360^0\\\\Hence,\\\\cos(\theta)=(√(21) )/(5)`

`\displaystyle\\a)\ sin(2\theta)=2sin(\theta)cos(\theta)\\\\sin(2\theta)=2(-(2)/(5))((√(21) )/(5))\\\\sin(2\theta)=-(4√(21) )/(25)`

`\displaystyle\\b)\ cos(2\theta)=cos^2(\theta)-sin^2(\theta)\\\\cos(2\theta)=((√(21) )/(5))^2-(-(2)/(5))^2 \\\\cos(2\theta)=(21)/(25)-(4)/(25) \\\\cos(2\theta)=(17)/(25)`