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Question
For any acute angle θ,cos(θ−3π)=
For any acute angle θ,cos(θ−3π)=
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Solution
The correct option is D −cosθ
Given an acute angle θ and we have to find the value of cos(θ−3π).
Now, we know for any angle ϕ,
cos(−ϕ)=cosϕ⋯(a)
Also, cos(θ−3π)=cos{−(3π−θ)}
Thus, using the condition (a):
cos{−(3π−θ)}=cos(3π−θ)
Since, θ is an acute angle, 3π−θ will correpsond to an angle in the 2nd quadrant.
cos(3π−θ)=−cosθ
Similarly, for sin(θ−3π)
From the relation: sin(−ϕ)=−sinϕ
sin(θ−3π)=−sin(3π−θ)
And, since 3π−θ corresponds to an angle in the second quadrant, where sin of any angle is positive.
⇒−sin(3π−θ)=−sinθ
Given an acute angle θ and we have to find the value of cos(θ−3π).
Now, we know for any angle ϕ,
cos(−ϕ)=cosϕ⋯(a)
Also, cos(θ−3π)=cos{−(3π−θ)}
Thus, using the condition (a):
cos{−(3π−θ)}=cos(3π−θ)
Since, θ is an acute angle, 3π−θ will correpsond to an angle in the 2nd quadrant.
cos(3π−θ)=−cosθ
Similarly, for sin(θ−3π)
From the relation: sin(−ϕ)=−sinϕ
sin(θ−3π)=−sin(3π−θ)
And, since 3π−θ corresponds to an angle in the second quadrant, where sin of any angle is positive.
⇒−sin(3π−θ)=−sinθ
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