Question

SUBJECT - Mathematics
CLASS - 10
TOPIC - Trigonometry
$\large{\sf{\underline{QUESTION:-}}}$
If 2 sin θ = 2 - cos θ and 0° < θ < 90°. Then, Find the value of sin θ.

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Answer 1

Step-by-step explanation:

topic :

• Trigonometry

Question :

• if 2 sin θ = 2 - cos θ and 0° < θ < 90°. then, find the value of sin θ.

given :

• sin θ = 2 - cos θ

• sin θ = 0° < θ < 90°.

to find :

• find the value of sin θ = ?

• find the value of sin θ = ?

solution :

Given 2 sin + cos θ = 2

• cos θ.= 2 - 2 sin

Substituting the value cos θ

• sin² θ+ (2 - 2 sin θ)² = 1

• 5 sin²θ -8 sin θ +3 = 0

• 5 sin θ -3=0

• sin θ = 3/5

• cos θ = 4/5

Putting the value in the equation

• sin θ - 2 cos θ = 3/5 - 2× 4/5 = -1

We cancel this value as

• 0° ≤0 ≤ 90°

• then, we come to other factors of equation

• sin θ - 1 = 0

• sin θ = 1

• cos θ = 0

• sin θ - 2 cos θ = 1 - 2 x 0 = 1

Answer 2

GIVEN :–

$\\ \implies \sf 2 \sin\theta= 2 - \cos \theta \: \: and \: \: 0^{\degree}< \theta< 90^{\degree}$

TO FIND :–

$\\ \implies \sf Value \: \: of \: \sin\theta=?$

SOLUTION :–

$\\ \implies \sf 2 \sin\theta= 2 - \cos \theta$

$\\ \implies \sf 2 \sin\theta - 2 + \cos \theta = 0$

$\\ \implies \sf 2 \sin\theta - 2 = - \cos \theta$

• Square on both sides –

$\\ \implies \sf (2 \sin\theta - 2) ^{2} = (- \cos \theta)^{2}$

$\\ \implies \sf (2 \sin\theta - 2) ^{2} = \cos^{2} \theta$

$\\ \implies \sf 4(\sin\theta - 1) ^{2} = \cos^{2} \theta$

$\\ \implies \sf 4(1-\sin\theta) ^{2} = \cos^{2} \theta$

$\\ \implies \sf 4(1 + \sin^{2} \theta - 2 \sin \theta) =1 - \sin^{2} \theta$

$\\ \implies \sf 4+4\sin^{2} \theta - 8\sin \theta =1 - \sin^{2} \theta$

$\\ \implies \sf 3+5\sin^{2} \theta - 8\sin \theta =0$

$\\ \implies \sf 5\sin^{2} \theta - 8\sin \theta + 3=0$

• By using quadratic formula –

$\\ \implies \sf \sin \theta = \dfrac{8 \pm \sqrt{( - 8)^{2} - 4(5)(3)} }{2 \times 5}$

$\\ \implies \sf \sin \theta = \dfrac{8 \pm \sqrt{64-60} }{10}$

$\\ \implies \sf \sin \theta = \dfrac{8 \pm \sqrt{4} }{10}$

$\\ \implies \sf \sin \theta = \dfrac{8 \pm 2}{10}$

$\\ \implies \sf \sin \theta = \dfrac{8 + 2}{10} ,\dfrac{8 - 2}{10}$

$\\ \implies \sf \sin \theta = \dfrac{10}{10} ,\dfrac{6}{10}$

$\\ \implies \sf \sin \theta = 1 ,\dfrac{3}{5}$

$\\ \: \red{ \:[ \: \: \: \because \sf \: \: \: 0^{\degree}< \theta< 90^{\degree} \: \: ]}$

$\\ \: \: \: \therefore \: \: \sf \sin \theta = 1 \: \: (x)$

• Hence –

$\\ \implies \pink{ \boxed{ \sf \sin \theta = \dfrac{3}{5}}}$

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$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$