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Math  /  Trigonometry

Question1. Simplify the expressions (a) sin⁡2Acos⁡A\frac{\sin 2 A}{\cos A}, (b) 2tan⁡Acos⁡2A2 \tan A \cos ^{2} A,
2. If cos⁡A=−23\cos A=-\frac{2}{3} find cos⁡2A\cos 2 A.
3. Angle BB is acute and tan⁡B=43\tan B=\frac{4}{3}. Find (a) sin⁡B\sin B (b) cos⁡B\cos B (c) sin⁡2B\sin 2 B
4. If sin⁡C=1/5\sin C=1 / \sqrt{5} find sin⁡2C,cos⁡2C\sin 2 C, \cos 2 C and tan⁡2C\tan 2 C if (a) CC is acute, (b) CC is obtuse.

Studdy Solution

STEP 1

1. We are asked to simplify trigonometric expressions and find specific trigonometric values.
2. We will use trigonometric identities to simplify and calculate the expressions.
3. For finding values, we will use known identities and given values.

STEP 2

1. Simplify the expressions (a) and (b).
2. Calculate cos⁡2A\cos 2A given cos⁡A=−23\cos A = -\frac{2}{3}.
3. Find sin⁡B\sin B, cos⁡B\cos B, and sin⁡2B\sin 2B given tan⁡B=43\tan B = \frac{4}{3}.
4. Calculate sin⁡2C\sin 2C, cos⁡2C\cos 2C, and tan⁡2C\tan 2C for both acute and obtuse angles given sin⁡C=15\sin C = \frac{1}{\sqrt{5}}.

STEP 3

(a) Simplify sin⁡2Acos⁡A\frac{\sin 2A}{\cos A}.
Use the double angle identity for sine: sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2 \sin A \cos A.
sin⁡2Acos⁡A=2sin⁡Acos⁡Acos⁡A=2sin⁡A\frac{\sin 2A}{\cos A} = \frac{2 \sin A \cos A}{\cos A} = 2 \sin A

STEP 4

(b) Simplify 2tan⁡Acos⁡2A2 \tan A \cos^2 A.
Use the identity tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}.
2tan⁡Acos⁡2A=2(sin⁡Acos⁡A)cos⁡2A=2sin⁡Acos⁡A2 \tan A \cos^2 A = 2 \left(\frac{\sin A}{\cos A}\right) \cos^2 A = 2 \sin A \cos A

STEP 5

Calculate cos⁡2A\cos 2A given cos⁡A=−23\cos A = -\frac{2}{3}.
Use the double angle identity for cosine: cos⁡2A=2cos⁡2A−1\cos 2A = 2 \cos^2 A - 1.
First, calculate cos⁡2A\cos^2 A:
cos⁡2A=(−23)2=49\cos^2 A = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}
Now, substitute into the identity:
cos⁡2A=2(49)−1=89−1=89−99=−19\cos 2A = 2 \left(\frac{4}{9}\right) - 1 = \frac{8}{9} - 1 = \frac{8}{9} - \frac{9}{9} = -\frac{1}{9}

STEP 6

Find sin⁡B\sin B and cos⁡B\cos B given tan⁡B=43\tan B = \frac{4}{3}.
Use the identity tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}.
Assume sin⁡B=4k\sin B = 4k and cos⁡B=3k\cos B = 3k.
Then tan⁡B=4k3k=43\tan B = \frac{4k}{3k} = \frac{4}{3}.
Use the Pythagorean identity sin⁡2B+cos⁡2B=1\sin^2 B + \cos^2 B = 1:
(4k)2+(3k)2=1  ⟹  16k2+9k2=1  ⟹  25k2=1  ⟹  k2=125  ⟹  k=15(4k)^2 + (3k)^2 = 1 \implies 16k^2 + 9k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5}
Thus, sin⁡B=4k=45\sin B = 4k = \frac{4}{5} and cos⁡B=3k=35\cos B = 3k = \frac{3}{5}.

STEP 7

Calculate sin⁡2B\sin 2B.
Use the double angle identity for sine: sin⁡2B=2sin⁡Bcos⁡B\sin 2B = 2 \sin B \cos B.
sin⁡2B=2(45)(35)=2×1225=2425\sin 2B = 2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right) = 2 \times \frac{12}{25} = \frac{24}{25}

STEP 8

Calculate sin⁡2C\sin 2C, cos⁡2C\cos 2C, and tan⁡2C\tan 2C given sin⁡C=15\sin C = \frac{1}{\sqrt{5}}.
First, find cos⁡C\cos C using sin⁡2C+cos⁡2C=1\sin^2 C + \cos^2 C = 1.
sin⁡2C=(15)2=15\sin^2 C = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}
cos⁡2C=1−15=45\cos^2 C = 1 - \frac{1}{5} = \frac{4}{5}
(a) If CC is acute, cos⁡C=45=25\cos C = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}.
(b) If CC is obtuse, cos⁡C=−45=−25\cos C = -\sqrt{\frac{4}{5}} = -\frac{2}{\sqrt{5}}.
Now calculate sin⁡2C\sin 2C and cos⁡2C\cos 2C:
sin⁡2C=2sin⁡Ccos⁡C=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}
cos⁡2C=2cos⁡2C−1\cos 2C = 2 \cos^2 C - 1
For acute CC:
cos⁡2C=2(25)2−1=2×45−1=85−1=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:
cos⁡2C=2(−25)2−1=2×45−1=85−1=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 tan⁡2C\tan 2C:
tan⁡2C=sin⁡2Ccos⁡2C=4535=43\tan 2C = \frac{\sin 2C}{\cos 2C} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}

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