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It's addma under trigonometry compound and multiple angles.show simply

Chemistry

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2 parts
a

−6365-\frac{63}{65}

b

6365\frac{63}{65}

Step-by-step solution

4 steps

Here's how to solve the problem step-by-step:

Given: In a triangle ABC, cos⁡A=1213\cos A = \frac{12}{13} and cos⁡B=−45\cos B = -\frac{4}{5}.

Part a· 2 steps

Calculate cos⁡(A+B)\cos(A+B)

  1. Find sin⁡A\sin A and sin⁡B\sin B.

    Since AA and BB are angles in a triangle, sin⁡A>0\sin A > 0 and sin⁡B>0\sin B > 0.
    Using the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1:
    For angle AA:

    sin⁡2A=1−cos⁡2A=1−(1213)2=1−144169=169−144169=25169\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} sin⁡A=25169=513\sin A = \sqrt{\frac{25}{169}} = \frac{5}{13}

    For angle BB:

    sin⁡2B=1−cos⁡2B=1−(−45)2=1−1625=25−1625=925\sin^2 B = 1 - \cos^2 B = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} sin⁡B=925=35\sin B = \sqrt{\frac{9}{25}} = \frac{3}{5}

  2. Apply the compound angle formula for cos⁡(A+B)\cos(A+B).

    The formula is cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A \cos B - \sin A \sin B.
    Substitute the known values:

    cos⁡(A+B)=(1213)(−45)−(513)(35)\cos(A+B) = \left(\frac{12}{13}\right) \left(-\frac{4}{5}\right) - \left(\frac{5}{13}\right) \left(\frac{3}{5}\right) cos⁡(A+B)=−4865−1565\cos(A+B) = -\frac{48}{65} - \frac{15}{65} cos⁡(A+B)=−48−1565\cos(A+B) = \frac{-48 - 15}{65} cos⁡(A+B)=−6365\cos(A+B) = \boxed{-\frac{63}{65}}

Part b

6365\frac{63}{65}

Calculate cos⁡C\cos C

  1. Use the angle sum property of a triangle.

    In a triangle, A+B+C=180∘A+B+C = 180^\circ.
    Therefore, C=180∘−(A+B)C = 180^\circ - (A+B).

  2. Find cos⁡C\cos C using the identity cos⁡(180∘−θ)=−cos⁡θ\cos(180^\circ - \theta) = -\cos \theta.

    cos⁡C=cos⁡(180∘−(A+B))\cos C = \cos(180^\circ - (A+B)) cos⁡C=−cos⁡(A+B)\cos C = -\cos(A+B)

    Substitute the value of cos⁡(A+B)\cos(A+B) from Part a):

    cos⁡C=−(−6365)\cos C = -\left(-\frac{63}{65}\right) cos⁡C=6365\cos C = \boxed{\frac{63}{65}}
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