11. Express root3 sin theta + cos theta as a sine of an angle
Answers
given : \sqrt{3}\sin \theta +\cos\theta
3
sinθ+cosθ
we have to express it in the form of sin only
therefore ,
\begin{gathered}\sqrt{3}\sin \theta +\cos\theta \\\\\text{multiply and divide by }\\\\\sqrt{(3)^2+(1)^2}\\\\we get \\\\(\frac{\sqrt{3}\sin \theta}{\sqrt{(3)^2+(1)^2}}+\frac{\cos\theta}{\sqrt{(3)^2+(1)^2}})\times \sqrt{(3)^2+(1)^2}\end{gathered}
3
sinθ+cosθ
multiply and divide by
(3)
2
+(1)
2
weget
(
(3)
2
+(1)
2
3
sinθ
+
(3)
2
+(1)
2
cosθ
)×
(3)
2
+(1)
2
\begin{gathered}(\frac{\sqrt{3}}{\sqrt{4}}\sin\theta +\frac{1}{\sqrt{4}}\cos\theta)\times \sqrt{4}\\\\2 ( \frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta )\\\\2 ( \sin\theta \cos \frac{\pi}{6}+\cos\theta \sin \frac{\pi}{6} )\\\\\sin A \cos B + \cos A \sin B = \sin (A+B)\\\\2\sin (\theta+\frac{\pi}{6})\end{gathered}
(
4
3
sinθ+
4
1
cosθ)×
4
2(
2
3
sinθ+
2
1
cosθ)
2(sinθcos
6
π
+cosθsin
6
π
)
sinAcosB+cosAsinB=sin(A+B)
2sin(θ+
6
π
)
hence ,
\sqrt{3}\sin \theta +\cos\theta = 2\sin (\theta+\frac{\pi}{6})
3
sinθ+cosθ=2sin(θ+
6
π
)