Trigonometry identities

Geometry Level 3

2 sin ⁡ x + 3 cos ⁡ x = 3 \large 2\sin x+3\cos x =3

If real number x ∈ ( 0 , π ) x \in (0, \pi) satisfies the equation above and that sin ⁡ x = a b \sin x = \dfrac ab and cos ⁡ x = c b \cos x = \dfrac cb , where a a , b b and c c are pairwise coprime positive integers . Find a + b + c a+b+c .


The answer is 30.

Djordje Veljkovic by Djordje Veljkovic , 19, Serbia

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2 solutions

2 sin ⁡ x + 3 cos ⁡ x = 3 Squaring both sides 4 sin ⁡ 2 x + 12 sin ⁡ x cos ⁡ x + 9 cos ⁡ 2 x = 9 4 sin ⁡ 2 x + 12 sin ⁡ x cos ⁡ x + 9 ( 1 − sin ⁡ 2 x ) = 9 12 sin ⁡ x cos ⁡ x + 9 − 5 sin ⁡ 2 x = 9 12 sin ⁡ x cos ⁡ x − 5 sin ⁡ 2 x = 0 12 cos ⁡ x = 5 sin ⁡ x ⟹ tan ⁡ x = 12 5 sin ⁡ x = 12 13 cos ⁡ x = 5 13 \begin{aligned} 2 \sin x + 3 \cos x & = 3 & \small \color{#3D99F6}{\text{Squaring both sides}} \\ 4 \sin^2 x + 12 \sin x \cos x + 9 \cos^2 x & = 9 \\ 4 \sin^2 x + 12 \sin x \cos x + 9 (1-\sin^2 x) & = 9 \\ 12 \sin x \cos x + 9 - 5 \sin^2 x & = 9 \\ 12 \sin x \cos x - 5 \sin^2 x & = 0 \\ 12 \cos x & = 5 \sin x \\ \implies \tan x & = \frac {12}5 \\ \sin x & = \frac {12}{13} \\ \cos x & = \frac 5{13} \end{aligned}

⟹ a + b + c = 12 + 13 + 5 = 30 \implies a + b + c = 12 + 13 + 5 = \boxed{30}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Djordje Veljkovic , I have edited your problem wording. Note that x = 0 x=0 is a solution, therefore, I have included x ∈ ( 0 , π ) x \in (0, \pi) which means that x ≠ 0 x \ne 0 . I have also changed x , y , z x, y, z to a , b , c a, b, c because you have used x x . You have to mentioned that a , b , c a,b,c are coprime integers, because 24 , 26 , 10 24, 26, 10 are also solutions.

Chew-Seong Cheong - 4 years, 9 months ago

Thank you, I always have problems wording problems in English. I knew something was wrong, I just didn't know what. Thank you!

Djordje Veljkovic - 4 years, 9 months ago

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You will learn.

Chew-Seong Cheong - 4 years, 9 months ago
Sabhrant Sachan
Sep 4, 2016

Put sin ⁡ x = 2 tan ⁡ x 2 1 + tan ⁡ 2 x 2 and cos ⁡ x = 1 − tan ⁡ 2 x 2 1 + tan ⁡ 2 x 2 Now, the Given equation is : 4 tan ⁡ x 2 1 + tan ⁡ 2 x 2 + 3 − 3 tan ⁡ 2 x 2 1 + tan ⁡ 2 x 2 = 3 Take tan ⁡ x 2 = X 4 X 1 + X 2 + 3 − 3 X 2 1 + X 2 = 3 4 X − 3 X 2 = 3 X 2 ⟹ X = 2 3 , 0 ⟹ sin ⁡ x = 4 3 1 + 4 9 = 12 13 , 0 cos ⁡ x = 5 13 , 1 Answer : 12 + 13 + 5 = 30 \text{Put } \sin{x} = \dfrac{2\tan{\frac{x}{2}}}{1+\tan^{2}{\frac{x}{2}}} \text{ and } \cos{x} = \dfrac{1-\tan^{2}{\frac{x}{2}}}{1+\tan^{2}{\frac{x}{2}}} \\ \text{Now, the Given equation is : } \\ \dfrac{4\tan{\frac{x}{2}}}{1+\tan^{2}{\frac{x}{2}}} + \dfrac{3-3\tan^{2}{\frac{x}{2}}}{1+\tan^{2}{\frac{x}{2}}}=3 \\ \text{Take } \tan{\frac{x}{2}}=X \\ \dfrac{4X}{1+X^2}+\dfrac{3-3X^2}{1+X^2}=3 \\ 4X-3X^2=3X^2 \\ \implies X=\frac{2}{3},0 \\ \implies \sin{x} = \dfrac{\frac43}{1+\frac49} = \dfrac{12}{13} , 0 \\ \quad \quad \quad \cos{x} = \dfrac{5}{13} , 1 \\ \text{Answer : } 12+13+5 = \boxed{30}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Your method is very good, although I was thinking about using basic trigonometric functions which would lead to a quadratic equation, but that is too complicated. Thanks for the contribution! :)

Djordje Veljkovic - 4 years, 9 months ago

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Your Welcome :)

Sabhrant Sachan - 4 years, 9 months ago

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