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Evaluate: 1. tan 160°, cos 300° 2. sin 340°, cos 150° 3. sin 130°, sin(-70°), cos 350°

Mathematics

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Step-by-step solution

7 steps

Right Triangle (opposite = 12, hypotenuse = 18)

  1. sin⁡θ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}.

    Substitute: sin⁡θ=1218\sin \theta = \frac{12}{18}.

    Simplify:

    sin⁡θ=23\sin \theta = \frac{2}{3}
  2. Adjacent side = hypotenuse2−opposite2\sqrt{\text{hypotenuse}^2 - \text{opposite}^2}.

    Substitute: Adjacent side = 182−122=324−144=180\sqrt{18^2 - 12^2} = \sqrt{324 - 144} = \sqrt{180}.

    Simplify:

    180=36×5=65\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}
  3. cos⁡θ=adjacenthypotenuse\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}.

    Substitute: cos⁡θ=6518\cos \theta = \frac{6\sqrt{5}}{18}.

    Simplify:

    cos⁡θ=53\cos \theta = \frac{\sqrt{5}}{3}
  4. tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}}.

    Substitute: tan⁡θ=1265\tan \theta = \frac{12}{6\sqrt{5}}.

    Simplify:

    tan⁡θ=25=255\tan \theta = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}

    Trigonometric Table (30°, 45°, 60°)

    Using 30°-60°-90° triangle (sides opposite: 1, 3\sqrt{3}, hypotenuse 2).

  5. sin⁡30∘=opp (30°)hyp\sin 30^\circ = \frac{\text{opp (30°)}}{\text{hyp}}.

    Substitute: sin⁡30∘=12\sin 30^\circ = \frac{1}{2}.

    sin⁡30∘=12\sin 30^\circ = \frac{1}{2}

    cos⁡30∘=adj (30°)hyp=32\cos 30^\circ = \frac{\text{adj (30°)}}{\text{hyp}} = \frac{\sqrt{3}}{2}.

    cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}

    tan⁡30∘=opp (30°)adj (30°)=13=33\tan 30^\circ = \frac{\text{opp (30°)}}{\text{adj (30°)}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.

    tan⁡30∘=33\tan 30^\circ = \frac{\sqrt{3}}{3}
  6. sin⁡60∘=opp (60°)hyp=32\sin 60^\circ = \frac{\text{opp (60°)}}{\text{hyp}} = \frac{\sqrt{3}}{2}.

    sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

    cos⁡60∘=adj (60°)hyp=12\cos 60^\circ = \frac{\text{adj (60°)}}{\text{hyp}} = \frac{1}{2}.

    cos⁡60∘=12\cos 60^\circ = \frac{1}{2}

    tan⁡60∘=opp (60°)adj (60°)=3\tan 60^\circ = \frac{\text{opp (60°)}}{\text{adj (60°)}} = \sqrt{3}.

    tan⁡60∘=3\tan 60^\circ = \sqrt{3}

    Using 45°-45°-90° triangle (sides opposite: 1, 1, hypotenuse 2\sqrt{2}).

  7. sin⁡45∘=opp (45°)hyp=12=22\sin 45^\circ = \frac{\text{opp (45°)}}{\text{hyp}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.

    sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2}

    cos⁡45∘=12=22\cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.

    cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2}

    tan⁡45∘=11=1\tan 45^\circ = \frac{1}{1} = 1.

    tan⁡45∘=1\tan 45^\circ = 1

    \sin \theta = 2/3, \cos \theta = \sqrt{5}/3, \tan \theta = 2\sqrt{5}/5

    30°: \sin=1/2, \cos=\sqrt{3}/2, \tan=\sqrt{3}/3

    45°: \sin=\sqrt{2}/2, \cos=\sqrt{2}/2, \tan=1

    60°: \sin=\sqrt{3}/2, \cos=1/2, \tan=\sqrt{3}

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