Question

if 3sin (theta) +4 cos (theta) =5 then find the value of 4 sin (theta) - 3 cos theta, also find sin theta and cos theta

Answer 1

3 sin x + 4 Cos x=5 3/5*sin x + 4/5 cos x =1 Cos 53° * sin x + sin 53° * cos x=1 Sin(53°+x)= Sin (90°) 53° + x = 90° x=37° 4 sin 37° - 3 cos 37°= 4*3/5-3*4/5 =0

Answer 2

Answer:

$4 sin\theta-3 cos\theta=0$

$sin\theta=\frac{3}{5} cos\theta=\frac{4}{5}$

Step-by-step explanation:

We are given that

$3 sin\theta+4 cos\theta=5$

We have to find the value of $4 sin\theta-3 cos\theta$ , $sin\theta$ and $cos\theta$

$\frac{3}{5}sin\theta+\frac{4}{5}cos\theta=1$

Comparing with the identity

$sin x cos y +cos x sin y=sin(x+y)$

Then we get $cos y =\frac{3}{5}$ and $siny =\frac{4}{5}$

$x= \theta$

$sin^{-1}\frac{4}{5}=53^{\circ}$ and $cos^{-1}\frac{3}{5}=53^{\circ}$

Substituting values in the given formula then we have

$sin(53^{\circ}+\theta)=1$

$53+\theta=sin^{-1}(1)$

We know that $sin^{-1}=90^{\circ}$

Therefore, $\theta+53=90$

$\theta=90-53=37^{\circ}$

$\theta=37^{\circ}$

The value of $4sin\theta-3cos\theta$

=$4\times sin37^{\circ}-3 cos37^{\circ}$

=$4\times \frac{3}{5}-3\times \frac{4}{5}$

=0

Hence, the value of $4sin\theta-3cos\theta$=0

$sin\theta=sin37^{\circ}=\frac{3}{5}$

$cos\theta=cos37^{\circ}=\frac{4}{5}$