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Q1. If cosA = 2/5, SinB = 12/13. Find a) Cos (A + B) b) Sin (A-B) c) Sin2A Q2. a) Solve the equation 2sin2 x − sinx −1 =0

Mathematics

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3 parts
a

a) cos⁡(A+B)=−3365\text{a) } \cos (A + B) = -\frac{33}{65}

b

b) sin⁡(A−B)=−1665\text{b) } \sin (A - B) = -\frac{16}{65}

c

c) sin⁡2A=2425\text{c) } \sin 2A = \frac{24}{25}

a) x=90∘,210∘,330∘\text{a) } x = 90^\circ, 210^\circ, 330^\circ

Step-by-step solution

14 steps

Here are the solutions to the structured questions:

Q1. Given cos⁡A=35\cos A = \frac{3}{5} and sin⁡B=1213\sin B = \frac{12}{13}.

First, we find the missing trigonometric values for angles A and B using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. Assuming A and B are acute angles (in the first quadrant):

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

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

So we have: sin⁡A=45\sin A = \frac{4}{5}, cos⁡A=35\cos A = \frac{3}{5}, sin⁡B=1213\sin B = \frac{12}{13}, cos⁡B=513\cos B = \frac{5}{13}.

Part a· 3 steps

Find cos⁡(A+B)\cos (A + B)

  1. Use the cosine addition formula

    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A \cos B - \sin A \sin B.

  2. Substitute the known values into the formula.

    cos⁡(A+B)=(35)(513)−(45)(1213)\cos (A + B) = \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) - \left(\frac{4}{5}\right)\left(\frac{12}{13}\right)

  3. Perform the multiplication and subtraction.

    cos⁡(A+B)=1565−4865=15−4865=−3365\cos (A + B) = \frac{15}{65} - \frac{48}{65} = \frac{15 - 48}{65} = \frac{-33}{65} a) cos⁡(A+B)=−3365\boxed{\text{a) } \cos (A + B) = -\frac{33}{65}}

Part b

b) sin⁡(A−B)=−1665\text{b) } \sin (A - B) = -\frac{16}{65}

Find sin⁡(A−B)\sin (A - B)

  1. Use the sine subtraction formula

    sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A \cos B - \cos A \sin B.

  2. Substitute the known values into the formula.

    sin⁡(A−B)=(45)(513)−(35)(1213)\sin (A - B) = \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{12}{13}\right)

  3. Perform the multiplication and subtraction.

    sin⁡(A−B)=2065−3665=20−3665=−1665\sin (A - B) = \frac{20}{65} - \frac{36}{65} = \frac{20 - 36}{65} = \frac{-16}{65} b) sin⁡(A−B)=−1665\boxed{\text{b) } \sin (A - B) = -\frac{16}{65}}

Part c

c) sin⁡2A=2425\text{c) } \sin 2A = \frac{24}{25}

Find sin⁡2A\sin 2A

  1. Use the sine double angle formula

    sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2 \sin A \cos A.

  2. Substitute the known values for sin⁡A\sin A and cos⁡A\cos A.

    sin⁡2A=2(45)(35)\sin 2A = 2 \left(\frac{4}{5}\right)\left(\frac{3}{5}\right)

  3. Perform the multiplication.

    sin⁡2A=2×1225=2425\sin 2A = 2 \times \frac{12}{25} = \frac{24}{25} c) sin⁡2A=2425\boxed{\text{c) } \sin 2A = \frac{24}{25}}

    Q2. a) Solve the equation 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0

  4. Let y=sin⁡xy = \sin x.

    The equation becomes a quadratic equation in terms of yy.

    2y2−y−1=02y^2 - y - 1 = 0

  5. Solve the quadratic equation for yy by factoring.

    (2y+1)(y−1)=0(2y + 1)(y - 1) = 0

  6. Set each factor to zero to find the possible values for yy.

    2y+1=0  ⟹  2y=−1  ⟹  y=−122y + 1 = 0 \implies 2y = -1 \implies y = -\frac{1}{2} y−1=0  ⟹  y=1y - 1 = 0 \implies y = 1

  7. Substitute back y=sin⁡xy = \sin x and find the values of xx in the range 0∘≤x<360∘0^\circ \le x < 360^\circ.

    For sin⁡x=1\sin x = 1:
    The angle xx where sin⁡x=1\sin x = 1 is 90∘90^\circ.

    For sin⁡x=−12\sin x = -\frac{1}{2}:
    The reference angle for sin⁡x=12\sin x = \frac{1}{2} is 30∘30^\circ. Since sin⁡x\sin x is negative, xx lies in the third and fourth quadrants.
    In the third quadrant: x=180∘+30∘=210∘x = 180^\circ + 30^\circ = 210^\circ.
    In the fourth quadrant: x=360∘−30∘=330∘x = 360^\circ - 30^\circ = 330^\circ.

  8. List all solutions for xx.

    x=90∘,210∘,330∘x = 90^\circ, 210^\circ, 330^\circ a) x=90∘,210∘,330∘\boxed{\text{a) } x = 90^\circ, 210^\circ, 330^\circ}
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