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प्रश्न
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
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उत्तर
4 cos2 A – 3 = 0
`=>` 4 cos2 A = 3
`=> cos^2A = 3/4`
`=> cosA = sqrt3/2`
We know cos 30° `= sqrt(3)/2`
So, A = 30°
L.H.S. = cos3 A = cos 90° = 0
R.H.S. = 4 cos3 A – 3 cos A
= 4 cos3 30° – 3 cos 30°
= `4(sqrt3/2)^3 - 3(sqrt3/2)`
= `(3sqrt3)/2 - (3sqrt3)/2`
= 0
L.H.S. = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
If `sin theta = x , " write the value of cot "theta .`
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If sin A = `1/2`, then the value of sec A is ______.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
