Problem 1 :
Prove that (i) cos(30° + x) = (√3 cos x − sin x)/2
(ii) cos(π + θ) = −cos θ
(iii) sin(π + θ) = −sin θ.
Solution :
(i) cos(30° + x) :
cos (A + B) = cos A cos B - sin A sin B
cos(30° + x) = cos 30 cos x - sin 30 sin x
= (√3/2) cos x - (1/2) sin x
= (√3 cos x - sin x)/2
(ii) cos(π + θ) :
(π + θ) lies in the 3rd quadrant
In third quadrant, we will have negative sign for all trigonometric ratios other than tan θ and cot θ. Here we have cos θ.
So, cos(π + θ) = -cos θ
(iii) sin(π + θ) = −sin θ :
(π + θ) lies in the 3rd quadrant
In third quadrant, we will have negative sign for all trigonometric ratios other than tan θ and cot θ. Here we have sin θ.
So, sin(π + θ) = -sin θ
Problem 2 :
Find a quadratic equation whose roots are sin 15° and cos 15°.
Solution :
Since the roots of the required quadratic equation are sin 15° and cos 15°.
let α = sin 15° and β = cos 15°
α = sin 15° = sin (45 - 30)
sin (A - B) = sin A cos B - cos A sin B
= sin 45 cos 30 - cos 45 sin 30
= (1/√2) (√3/2) - (1/√2) (1/2)
= (√3/2√2) - (1/2√2)
α = sin 15° = (√3 - 1)/2√2 ---(1)
α = cos 15° = cos (45 - 30)
cos (A - B) = cos A cos B + sin A sin B
= cos 45 cos 30 + sin 45 sin 30
= (1/√2) (√3/2) + (1/√2) (1/2)
= (√3/2√2) + (1/2√2)
β = cos 15° = (√3 + 1)/2√2 ---(2)
α + β |
αβ |
General form of quadratic equation
x2 + (α + β)x + α β = 0
x2 + (√3/√2)x + (1/4) = 0
4x2+ 2√3x + 1 = 0
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