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Sum of Trigonometric Ratios in Terms of Their Product

In a triangle...

Question

In a triangle $A B C$ , angle $A$ is greater than angle $B$ . If the measures of angles $A$ and $B$ satisfy the equation $3 sin x - 4 {sin}^{3} x - k = 0, 0 < k < 1$ , then the measure of angle $C$ is

\frac{π}{4}

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\frac{π}{2}

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\frac{2 π}{3}

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\frac{5 π}{6}

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Solution

The correct option is C $\frac{2 π}{3}$
$3 sin x - 4 {sin}^{3} x = k \Rightarrow sin 3 x = k$

As $A$ and $B$ satisfy this, we get

$sin 3 A = k$ and $sin 3 B = k$

$\Rightarrow sin 3 A = sin 3 B$

$\Rightarrow sin 3 A - sin 3 B = 0$

$\Rightarrow 2 cos (\frac{3 A + 3 B}{2}) sin (\frac{3 A - 3 B}{2}) = 0$

Using $A + B + C = π \Rightarrow A + B = π - C$

$cos (3 (\frac{π - C}{2})) sin (\frac{3 A - 3 B}{2}) = 0$

Now as $A > B \Rightarrow sin (\frac{3 A - 3 B}{2}) > 0$

$sin (\frac{3 C}{2}) = 0 \Rightarrow \frac{3 C}{2} = n π \Rightarrow C = \frac{2 n π}{3}$

For $n = 1$

$C = \frac{2 π}{3}$

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In a triangle ABC, angle A is greater than angle B. If the measures of angles A and B satisfy the equation 3sinx−4sin3x−k=0,0<k<1, then the measure of angle C is

The correct option is C 2π33sinx−4sin3x=k⇒sin3x=kAs A and B satisfy this, we getsin3A=k and sin3B=k⇒sin3A=sin3B⇒sin3A−sin3B=0⇒2cos(3A+3B2)sin(3A−3B2)=0Using A+B+C=π⇒A+B=π−Ccos(3(π−C2))sin(3A−3B2)=0Now as A>B⇒sin(3A−3B2)>0sin(3C2)=0⇒3C2=nπ⇒C=2nπ3For n=1C=2π3

In a triangle $A B C$ , angle $A$ is greater than angle $B$ . If the measures of angles $A$ and $B$ satisfy the equation $3 sin x - 4 {sin}^{3} x - k = 0, 0 < k < 1$ , then the measure of angle $C$ is