find the value of sin if 6cos^2=4 1÷2
Answers
Answer:
The value is \sin\theta=\pm \frac{1}{2}sinθ=±
2
1
.
Step-by-step explanation:
Given : Expression 6\cos^2\theta=4\frac{1}{2}6cos
2
θ=4
2
1
To find : The value of \sin \thetasinθ ?
Solution :
Re-write the expression as,
6\cos^2\theta=\frac{9}{2}6cos
2
θ=
2
9
\cos^2\theta=\frac{9}{2\times 6}cos
2
θ=
2×6
9
\cos^2\theta=\frac{3}{4}cos
2
θ=
4
3
Using trigonometric formula,
\sin^2\theta=1-\cos^2 \thetasin
2
θ=1−cos
2
θ
\sin^2\theta=1-\frac{3}{4}sin
2
θ=1−
4
3
\sin^2\theta=\frac{4-3}{4}sin
2
θ=
4
4−3
\sin^2\theta=\frac{1}{4}sin
2
θ=
4
1
Taking root both side,
\sin\theta=\sqrt{\frac{1}{4}}sinθ=
4
1
\sin\theta=\pm \frac{1}{2}sinθ=±
2
1
Therefore, the value is \sin\theta=\pm \frac{1}{2}sinθ=±
2
1
.
#Learn more
Sin^theta - 1 / 2 sin theta = 0 . Find the value of theta
https://brainly.in/question/760266
Answer:
sin theta = 1/2
Step-by-step explanation:
Refer Image
Thanks
Please Rate My Answer As Brainliest
