Answer on Question #63977 – Math – Trigonometry
Question
Find the value of
sin 4 A − cos 4 A = sin 2 A − cos 2 A \sin 4\mathrm{A} - \cos 4\mathrm{A} = \sin 2\mathrm{A} - \cos 2\mathrm{A} sin 4 A − cos 4 A = sin 2 A − cos 2 A
Solution
Using formulae
sin x − sin y = 2 sin x − y 2 ⋅ cos x + y 2 ; \sin x - \sin y = 2 \sin \frac{x - y}{2} \cdot \cos \frac{x + y}{2}; sin x − sin y = 2 sin 2 x − y ⋅ cos 2 x + y ; cos x − cos y = − 2 sin x + y 2 ⋅ sin x − y 2 ; \cos x - \cos y = -2 \sin \frac{x + y}{2} \cdot \sin \frac{x - y}{2}; cos x − cos y = − 2 sin 2 x + y ⋅ sin 2 x − y ;
compute
sin 4 A − cos 4 A = sin 2 A − cos 2 A ; \sin 4\mathrm{A} - \cos 4\mathrm{A} = \sin 2\mathrm{A} - \cos 2\mathrm{A}; sin 4 A − cos 4 A = sin 2 A − cos 2 A ; sin 4 A − sin 2 A = cos 4 A − cos 2 A ; \sin 4\mathrm{A} - \sin 2\mathrm{A} = \cos 4\mathrm{A} - \cos 2\mathrm{A}; sin 4 A − sin 2 A = cos 4 A − cos 2 A ; 2 sin A cos 3 A = − 2 sin 3 A ⋅ sin A ; 2 \sin A \cos 3A = -2 \sin 3A \cdot \sin A; 2 sin A cos 3 A = − 2 sin 3 A ⋅ sin A ; cos 3 A = − sin 3 A or sin A = 0. \cos 3\mathrm{A} = -\sin 3\mathrm{A} \text{ or } \sin A = 0. cos 3 A = − sin 3 A or sin A = 0.
If cos 3 A = − sin 3 A \cos 3\mathrm{A} = -\sin 3\mathrm{A} cos 3 A = − sin 3 A , then
cos 3 A + sin 3 A = 0 ; \cos 3\mathrm{A} + \sin 3\mathrm{A} = 0; cos 3 A + sin 3 A = 0 ; 2 sin ( 3 A + π / 4 ) = 0 ; \sqrt{2} \sin (3\mathrm{A} + \pi/4) = 0; 2 sin ( 3 A + π /4 ) = 0 ; sin ( 3 A + π / 4 ) = 0 ; \sin (3\mathrm{A} + \pi/4) = 0; sin ( 3 A + π /4 ) = 0 ; 3 A + π / 4 = k π , k ∈ Z 3\mathrm{A} + \pi/4 = k\pi, k \in \mathbb{Z} 3 A + π /4 = kπ , k ∈ Z A = − π / 12 + k π / 3 , k ∈ Z . A = -\pi/12 + k\pi/3, k \in \mathbb{Z}. A = − π /12 + kπ /3 , k ∈ Z .
If sin A = 0 \sin A = 0 sin A = 0 , then A = m π A = m\pi A = mπ , m ∈ Z m \in \mathbb{Z} m ∈ Z .
Thus, A = − π / 12 + k π / 3 A = -\pi/12 + k\pi/3 A = − π /12 + kπ /3 , k ∈ Z k \in \mathbb{Z} k ∈ Z or A = m π A = m\pi A = mπ , m ∈ Z m \in \mathbb{Z} m ∈ Z .
Answer: A = − π / 12 + k π / 3 A = -\pi/12 + k\pi/3 A = − π /12 + kπ /3 , k ∈ Z k \in \mathbb{Z} k ∈ Z ; A = m π A = m\pi A = mπ , m ∈ Z m \in \mathbb{Z} m ∈ Z .
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