Solve - -3 - 1 cos - - -3 - 1 sin - - 2

(√(3) 1)cosθ +(√(3)+1)sinθ =2 → (1) Put √(3) 1=rsinα and √(3)+1=rcosα r^2=(√(3)+1)^2+(√(3)+1)^2⇒ r^2=8⇒ r=2√(2) Also tanα =√(3) ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Solve : √3 ...

Question

Solve : $(\sqrt{3} - 1) c o s θ + (\sqrt{3} + 1) s i n θ = 2$

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Solution

$(\sqrt{3} - 1) cos θ + (\sqrt{3} + 1) sin θ = 2$ $\to (1)$
Put $\sqrt{3} - 1 = r sin α$ and $\sqrt{3} + 1 = r cos α$
$r^{2} = (\sqrt{3} + 1)^{2} + (\sqrt{3} + 1)^{2} \Rightarrow r^{2} = 8 \Rightarrow r = 2 \sqrt{2}$
Also $tan α = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \Rightarrow α = \frac{π}{12}$
From $(1)$ we have
$r sin α cos θ + r cos α sin θ = 2$
$r sin (θ + α) = 2$
$sin (θ + α) = \frac{1}{\sqrt{2}} \cdot sin (θ + α) = sin \frac{π}{4}$
$θ + α = n π + (- 1)^{n} \frac{π}{4}$
$θ = n π + (- 1)^{n} \frac{π}{4} - \frac{π}{12}$ $n \in Z$ .

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Similar questions

$Solve the following equations : (i) s i n θ + c o s θ = \sqrt{2} (i i) \sqrt{3} c o s θ + s i n θ = 1 (i i i) s i n θ + c o s θ = 1 (i v) c o s e c θ = 1 + c o s θ (v) (\sqrt{3} - 1) c o s θ + (\sqrt{3} + 1) s i n θ = 2$

Q. General solution of the equation
(i)

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

Q. If

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2,

then for

n \in I, θ =

Q. General solution of the equation

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

Q. For

n \in z

the general solution of

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

(n ϵ z)

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Solve : (√3−1)cosθ+(√3+1)sinθ=2

(√3−1)cosθ+(√3+1)sinθ=2 →(1)Put √3−1=rsinα and √3+1=rcosαr2=(√3+1)2+(√3+1)2⇒r2=8⇒r=2√2Also tanα=√3−1√3+1=2−√3⇒α=π12From (1) we haversinαcosθ+rcosαsinθ=2rsin(θ+α)=2sin(θ+α)=1√2⋅sin(θ+α)=sinπ4θ+α=nπ+(−1)nπ4θ=nπ+(−1)nπ4−π12 n∈Z.

Solve : $(\sqrt{3} - 1) c o s θ + (\sqrt{3} + 1) s i n θ = 2$