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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 identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Eliminate θ if x = r cosθ and y = r sinθ.
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`
