Answer on Question #52820 – Math – Trigonometry
Verify that the following equality is an identity:
sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α ) = 1 4 sin ( 3 α ) \sin \alpha \cdot \sin (60{}^\circ - \alpha) \cdot \sin (60{}^\circ + \alpha) = \frac{1}{4} \sin (3\alpha) sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α ) = 4 1 sin ( 3 α )
Solution:
Let's consider the left side of the equality z l = sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α ) z_l = \sin \alpha \cdot \sin (60{}^\circ - \alpha) \cdot \sin (60{}^\circ + \alpha) z l = sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α )
First let's simplify z l z_l z l
sin ( 60 ∘ − α ) = sin ( 60 ∘ ) ⋅ cos α − cos ( 60 ∘ ) ⋅ sin α = 3 2 cos α − 1 2 sin α \sin (60{}^\circ - \alpha) = \sin (60{}^\circ) \cdot \cos \alpha - \cos (60{}^\circ) \cdot \sin \alpha = \frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha sin ( 60 ∘ − α ) = sin ( 60 ∘ ) ⋅ cos α − cos ( 60 ∘ ) ⋅ sin α = 2 3 cos α − 2 1 sin α sin ( 60 ∘ + α ) = sin ( 60 ∘ ) ⋅ cos α + cos ( 60 ∘ ) ⋅ sin α = 3 2 cos α + 1 2 sin α \sin (60{}^\circ + \alpha) = \sin (60{}^\circ) \cdot \cos \alpha + \cos (60{}^\circ) \cdot \sin \alpha = \frac{\sqrt{3}}{2} \cos \alpha + \frac{1}{2} \sin \alpha sin ( 60 ∘ + α ) = sin ( 60 ∘ ) ⋅ cos α + cos ( 60 ∘ ) ⋅ sin α = 2 3 cos α + 2 1 sin α z l = sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α ) = = sin α ⋅ ( 3 2 cos α − 1 2 sin α ) ⋅ ( 3 2 cos α − 1 2 sin α ) = = sin α ⋅ ( 3 4 cos 2 α − 1 4 sin 2 α ) = 1 4 sin α ⋅ ( 3 cos 2 α − sin 2 α ) = = 1 4 sin α ⋅ ( 3 ⋅ cos 2 α ⏟ 1 − sin 2 α − sin 2 α ) = 1 4 sin α ⋅ ( 3 ⋅ ( 1 − sin 2 α ) − sin 2 α ) = = 1 4 sin α ⋅ ( 3 − 3 sin 2 α − sin 2 α ) = 1 4 ( 3 sin α − 4 sin 3 α ) \begin{array}{l}
z_l = \sin \alpha \cdot \sin (60{}^\circ - \alpha) \cdot \sin (60{}^\circ + \alpha) = \\
= \sin \alpha \cdot \left( \frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha \right) \cdot \left( \frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha \right) = \\
= \sin \alpha \cdot \left( \frac{3}{4} \cos^2 \alpha - \frac{1}{4} \sin^2 \alpha \right) = \frac{1}{4} \sin \alpha \cdot \left( 3 \cos^2 \alpha - \sin^2 \alpha \right) = \\
= \frac{1}{4} \sin \alpha \cdot \left( 3 \cdot \underbrace{\cos^2 \alpha}_{1 - \sin^2 \alpha} - \sin^2 \alpha \right) = \frac{1}{4} \sin \alpha \cdot \left( 3 \cdot (1 - \sin^2 \alpha) - \sin^2 \alpha \right) = \\
= \frac{1}{4} \sin \alpha \cdot \left( 3 - 3 \sin^2 \alpha - \sin^2 \alpha \right) = \frac{1}{4} (3 \sin \alpha - 4 \sin^3 \alpha)
\end{array} z l = sin α ⋅ sin ( 60 ∘ − α ) ⋅ sin ( 60 ∘ + α ) = = sin α ⋅ ( 2 3 cos α − 2 1 sin α ) ⋅ ( 2 3 cos α − 2 1 sin α ) = = sin α ⋅ ( 4 3 cos 2 α − 4 1 sin 2 α ) = 4 1 sin α ⋅ ( 3 cos 2 α − sin 2 α ) = = 4 1 sin α ⋅ ⎝ ⎛ 3 ⋅ 1 − s i n 2 α cos 2 α − sin 2 α ⎠ ⎞ = 4 1 sin α ⋅ ( 3 ⋅ ( 1 − sin 2 α ) − sin 2 α ) = = 4 1 sin α ⋅ ( 3 − 3 sin 2 α − sin 2 α ) = 4 1 ( 3 sin α − 4 sin 3 α )
Next, let's consider the right side of the equality z r = 1 4 sin 3 α z_r = \frac{1}{4} \sin 3\alpha z r = 4 1 sin 3 α
z r = 1 4 sin ( 3 α ) = 1 4 sin ( α + 2 α ) = = 1 4 ( sin α cos 2 α + cos α sin 2 α ) = ∣ cos 2 α = cos 2 α − sin 2 α = 1 − 2 sin 2 α sin 2 α = 2 sin α cos α ∣ = = 1 4 ( sin α ⋅ ( 1 − 2 sin 2 α ) + 2 cos 2 α sin α ) = 1 4 ( sin α ⋅ ( 1 − 2 sin 2 α ) + 2 ( 1 − sin 2 α ) sin α ) = = 1 4 ( sin α − 2 sin 3 α + 2 sin α ⋅ ( 1 − sin 2 α ) ) = 1 4 ( sin α − 2 sin 3 α + 2 sin α − 2 sin 3 α ) = = 1 4 ( 3 sin α − 4 sin 3 α ) \begin{array}{l}
z_r = \frac{1}{4} \sin (3\alpha) = \frac{1}{4} \sin (\alpha + 2\alpha) = \\
= \frac{1}{4} (\sin \alpha \cos 2\alpha + \cos \alpha \sin 2\alpha) = \left| \begin{array}{c} \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2 \sin^2 \alpha \\ \sin 2\alpha = 2 \sin \alpha \cos \alpha \end{array} \right| = \\
= \frac{1}{4} (\sin \alpha \cdot (1 - 2 \sin^2 \alpha) + 2 \cos^2 \alpha \sin \alpha) = \frac{1}{4} (\sin \alpha \cdot (1 - 2 \sin^2 \alpha) + 2 (1 - \sin^2 \alpha) \sin \alpha) = \\
= \frac{1}{4} (\sin \alpha - 2 \sin^3 \alpha + 2 \sin \alpha \cdot (1 - \sin^2 \alpha)) = \frac{1}{4} (\sin \alpha - 2 \sin^3 \alpha + 2 \sin \alpha - 2 \sin^3 \alpha) = \\
= \frac{1}{4} (3 \sin \alpha - 4 \sin^3 \alpha)
\end{array} z r = 4 1 sin ( 3 α ) = 4 1 sin ( α + 2 α ) = = 4 1 ( sin α cos 2 α + cos α sin 2 α ) = ∣ ∣ cos 2 α = cos 2 α − sin 2 α = 1 − 2 sin 2 α sin 2 α = 2 sin α cos α ∣ ∣ = = 4 1 ( sin α ⋅ ( 1 − 2 sin 2 α ) + 2 cos 2 α sin α ) = 4 1 ( sin α ⋅ ( 1 − 2 sin 2 α ) + 2 ( 1 − sin 2 α ) sin α ) = = 4 1 ( sin α − 2 sin 3 α + 2 sin α ⋅ ( 1 − sin 2 α )) = 4 1 ( sin α − 2 sin 3 α + 2 sin α − 2 sin 3 α ) = = 4 1 ( 3 sin α − 4 sin 3 α )
The given equality (1) is an identity, because the left side z l z_l z l z r z_r z r α \alpha α
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#339914 on Dec 2023