Math  /  Trigonometry

Question1. Simplify the expressions (a) sin2AcosA\frac{\sin 2 A}{\cos A}, (b) 2tanAcos2A2 \tan A \cos ^{2} A,
2. If cosA=23\cos A=-\frac{2}{3} find cos2A\cos 2 A.
3. Angle BB is acute and tanB=43\tan B=\frac{4}{3}. Find (a) sinB\sin B (b) cosB\cos B (c) sin2B\sin 2 B
4. If sinC=1/5\sin C=1 / \sqrt{5} find sin2C,cos2C\sin 2 C, \cos 2 C and tan2C\tan 2 C if (a) CC is acute, (b) CC is obtuse.

Studdy Solution
Calculate sin2C\sin 2C, cos2C\cos 2C, and tan2C\tan 2C given sinC=15\sin C = \frac{1}{\sqrt{5}}.
First, find cosC\cos C using sin2C+cos2C=1\sin^2 C + \cos^2 C = 1.
sin2C=(15)2=15\sin^2 C = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}
cos2C=115=45\cos^2 C = 1 - \frac{1}{5} = \frac{4}{5}
(a) If CC is acute, cosC=45=25\cos C = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}.
(b) If CC is obtuse, cosC=45=25\cos C = -\sqrt{\frac{4}{5}} = -\frac{2}{\sqrt{5}}.
Now calculate sin2C\sin 2C and cos2C\cos 2C:
sin2C=2sinCcosC=2×15×25=45\sin 2C = 2 \sin C \cos C = 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} = \frac{4}{5}
cos2C=2cos2C1\cos 2C = 2 \cos^2 C - 1
For acute CC:
cos2C=2(25)21=2×451=851=35\cos 2C = 2 \left(\frac{2}{\sqrt{5}}\right)^2 - 1 = 2 \times \frac{4}{5} - 1 = \frac{8}{5} - 1 = \frac{3}{5}
For obtuse CC:
cos2C=2(25)21=2×451=851=35\cos 2C = 2 \left(-\frac{2}{\sqrt{5}}\right)^2 - 1 = 2 \times \frac{4}{5} - 1 = \frac{8}{5} - 1 = \frac{3}{5}
Calculate tan2C\tan 2C:
tan2C=sin2Ccos2C=4535=43\tan 2C = \frac{\sin 2C}{\cos 2C} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}

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