Question

Cos theta plus sin theta equals root 2 cos theta

Find: cos theta minus sin theta

Answer 1

Answer:

The value of $Cos\theta - Sin\theta$ is $\sqrt{2}Sin\theta$.

Step-by-step explanation:

It is provided that: $Cos\theta + Sin\theta=\sqrt{2} Cos\theta$.

Square both sides of the above equation and solve as follows:

$(Cos\theta + Sin\theta)^{2}=(\sqrt{2} Cos\theta)^{2}\\Cos^{2}\theta+Sin^{2}\theta+2Cos\theta Sin\theta=2Cos^{2}\theta\\2Cos\theta Sin\theta=Cos^{2}\theta-Sin^{2}\theta$

Now compute the value of $Cos\theta - Sin\theta$ as follows:

$(Cos\theta - Sin\theta)^{2}=Cos^{2}\theta+Sin^{2}\theta-2Cos\theta Sin\theta\\=Cos^{2}\theta+Sin^{2}\theta-(Cos^{2}\theta-Sin^{2}\theta)\\=Cos^{2}\theta+Sin^{2}\theta-Cos^{2}\theta+Sin^{2}\theta\\=2Sin^{2}\theta\\(Cos\theta - Sin\theta)=\sqrt{2Sin^{2}\theta} \\=\sqrt{2}Sin\theta$

Thus, the value of $Cos\theta - Sin\theta$ is $\sqrt{2}Sin\theta$.

Answer 2

Answer: (1) √2 cos θ

Solution:

Given,

cos θ – sin θ = √2 sin θ

Squaring on both sides,

cos2θ + sin2θ – 2 sin θ cos θ = 2 sin2θ

1 – 2 sin θ cos θ = 2 sin2θ

2 sin θ cos θ = 1 – 2 sin2θ(i)

Now,

cos θ + sin θ

Squaring this expression,

(cos θ + sin θ)2 = cos2θ + sin2θ + 2 sin θ cos θ

= 1 + 1 – 2 sin2θ {from (i)}

= 2 – 2 sin2θ

= 2(1 – sin2θ)

= 2 cos2θ

Therefore, cos θ + sin θ = √2 cos θ