Question #63977

Find the value of:-
sin4A-cos4A=sin2A-cos2A
1

Expert's answer

2016-12-12T11:25:11-0500

Answer on Question #63977 – Math – Trigonometry

Question

Find the value of

sin4Acos4A=sin2Acos2A\sin 4\mathrm{A} - \cos 4\mathrm{A} = \sin 2\mathrm{A} - \cos 2\mathrm{A}

Solution

Using formulae


sinxsiny=2sinxy2cosx+y2;\sin x - \sin y = 2 \sin \frac{x - y}{2} \cdot \cos \frac{x + y}{2};cosxcosy=2sinx+y2sinxy2;\cos x - \cos y = -2 \sin \frac{x + y}{2} \cdot \sin \frac{x - y}{2};


compute


sin4Acos4A=sin2Acos2A;\sin 4\mathrm{A} - \cos 4\mathrm{A} = \sin 2\mathrm{A} - \cos 2\mathrm{A};sin4Asin2A=cos4Acos2A;\sin 4\mathrm{A} - \sin 2\mathrm{A} = \cos 4\mathrm{A} - \cos 2\mathrm{A};2sinAcos3A=2sin3AsinA;2 \sin A \cos 3A = -2 \sin 3A \cdot \sin A;cos3A=sin3A or sinA=0.\cos 3\mathrm{A} = -\sin 3\mathrm{A} \text{ or } \sin A = 0.


If cos3A=sin3A\cos 3\mathrm{A} = -\sin 3\mathrm{A}, then


cos3A+sin3A=0;\cos 3\mathrm{A} + \sin 3\mathrm{A} = 0;2sin(3A+π/4)=0;\sqrt{2} \sin (3\mathrm{A} + \pi/4) = 0;sin(3A+π/4)=0;\sin (3\mathrm{A} + \pi/4) = 0;3A+π/4=kπ,kZ3\mathrm{A} + \pi/4 = k\pi, k \in \mathbb{Z}A=π/12+kπ/3,kZ.A = -\pi/12 + k\pi/3, k \in \mathbb{Z}.


If sinA=0\sin A = 0, then A=mπA = m\pi, mZm \in \mathbb{Z}.

Thus, A=π/12+kπ/3A = -\pi/12 + k\pi/3, kZk \in \mathbb{Z} or A=mπA = m\pi, mZm \in \mathbb{Z}.

Answer: A=π/12+kπ/3A = -\pi/12 + k\pi/3, kZk \in \mathbb{Z}; A=mπA = m\pi, mZm \in \mathbb{Z}.

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