Question

cos theta + sin theta - sin 2theta = 1/2​

Answer 1

Answer:

θ = $\frac{n\pi}{2}\ +\ (-1)^{n}(\frac{-\pi}{4})$

Step-by-step explanation:

In this question,

We have been given that

Cosθ + sinθ -sin2θ = $\frac{1}{2}$

We need to find the value of θ

Cosθ + sinθ = $\frac{1}{2}\ +\ Sin2\theta$

Squaring both the sides we get,

$(Cos\theta)\ +\ Sin\theta)^{2}\ =\ (\frac{1}{2}\ +\ Sin2\theta)^{2}$

$Cos^{2}\theta\ +\ Sin^{2}\theta\ +\ 2Sin\thetaCos\theta\ =\ \frac{(1\ +\ 2Sin2\theta)^{2}}{4}$

Since $Cos^{2}\theta\ +\ Sin^{2}\theta\ =\ 1$

Therefore,

1 + $2Sin\theta Cos\theta\ =\ \frac{(1\ +\ 2Sin2\theta)^{2}}{4}$

Since$2Sin\theta Cos\theta\ =\ Sin2\theta$

$1\ +\ Sin2\theta\ =\ \frac{(1\ +\ 2Sin2\theta)^{2}}{4}$

Let 1 + Sin2θ = x

Then, $x\ =\ \frac{x^{2}}{4}$

On solving we get x as 4 , 0

Therefore 1 + sin2θ = 0,4

Sin2θ = -1,3

We know that,

if Sinθ = sinφ

Then θ =$n\pi\ +\ (-1)^{n}\o$

Sin2θ = -1

Sin2θ = $Sin\frac{-\pi}{2}$

Therefore using formula we get,

2θ = $n\pi\ +\ (-1)^{n}\frac{-\pi}{2}$

θ =  $\frac{n\pi}{2}\ +\ (-1)^{n}\frac{-\pi}{4}$

Answer 2

Answer:

Step-by-step explanation: