Question

cos theta minus sin theta as a cosine of an angle​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Shailesh}}$

(cosθ - sinθ) = √2(1/√2 . cosθ - 1/√2. sinθ)

we know, cos45° = 1/√2

and sin45° = 1/√2

= 2(cos45° . cosθ - sin45° . sinθ)

we know, formula cos(A + B) = cosA.cosB - sinA. sinB

so, cos45° . cosθ - sin45° . sinθ = cos(45° + θ)

hence, (cosθ - sinθ) = √2cos(45° + θ)

$\bf\huge\underline\red{Follow me}$

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

(cosθ - sinθ) = √2(1/√2 . cosθ - 1/√2. sinθ)

we know, cos45° = 1/√2

and sin45° = 1/√2

= 2(cos45° . cosθ - sin45° . sinθ)

we know, formula cos(A + B) = cosA.cosB - sinA. sinB

so, cos45° . cosθ - sin45° . sinθ = cos(45° + θ)

hence, (cosθ - sinθ) = √2cos(45° + θ)