Hint: tan⁡ϕ=ab sin⁡ϕa=cos⁡ϕb=±?a2+b2 Now replace the values of a,b

tan⁡ϕ=abThus,cos⁡ϕ=ba2+b2sin⁡ϕ=aa2+b2L.H.S.acos⁡θ+bsin⁡θ=a2+b2(acos⁡θ+bsin⁡θ)a2+b2=a2+b2(aa2+b2cos⁡θ+ba2+b2sin⁡θ)

Hint: tan⁡ϕ=ab sin⁡ϕa=cos⁡ϕb=±?a2+b2 Now replace the values of a,b

tan⁡ϕ=abThus,cos⁡ϕ=ba2+b2sin⁡ϕ=aa2+b2L.H.S.acos⁡θ+bsin⁡θ=a2+b2(acos⁡θ+bsin⁡θ)a2+b2=a2+b2(aa2+b2cos⁡θ+ba2+b2sin⁡θ)

"Ive" was answered by our community

Secondary
Geometry
Trigonometry
Trigonometric equations and identities

regatamin2

Answered question

2022-01-03

Ive

Answer & Explanation

John Koga Beginner2022-01-04Added 33 answers

Hint:

\tan ϕ = \frac{a}{b}

\frac{\sin ϕ}{a} = \frac{\cos ϕ}{b} = \pm \sqrt{\frac{?}{a^{2} + b^{2}}}

Now replace the values of a,b

censoratojk

Beginner2022-01-05Added 46 answers

HINT
Begin by assuming that

a \cos θ + b \sin θ

can be written in the form

R \sin (x + ϕ)

for some constants R and

ϕ

. Expand the second expression using the angle addition formula, form some equations, and you should have it.

Vasquez

Skilled2022-01-08Added 457 answers

$\begin{matrix} \tan ϕ = \frac{a}{b} \\ T h u s, \\ \cos ϕ = \frac{b}{\sqrt{a^{2} + b^{2}}} \\ \sin ϕ = \frac{a}{\sqrt{a^{2} + b^{2}}} \\ L . H . S . \\ a \cos θ + b \sin θ = \sqrt{a^{2} + b^{2}} \frac{(a \cos θ + b \sin θ)}{\sqrt{a^{2} + b^{2}}} \\ = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} \cos θ + \frac{b}{\sqrt{a^{2} + b^{2}}} \sin θ) \end{matrix}$

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