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Sign of Trigonometric Ratios in Different Quadrants

cos3A-cosAsin...

Question

$\frac{c o s 3 A - c o s A}{s i n 3 A - s i n A} + \frac{c o s 2 A - c o s 4 A}{s i n 4 A - s i n 2 A} =$ $\frac{sin A}{cos 2 A cos 3 A}$ . Is it true?If true enter 1 else 0.

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Solution

Given

$\frac{cos 3 A - cos A}{sin 3 A - sin A} + \frac{cos 2 A - cos 4 A}{sin 4 A - sin 2 A} = \frac{cos 2 A}{sin A cos 3 A}$

$\frac{cos 3 A - cos A}{sin 3 A - sin A}$

$⟹ \frac{- 2 sin 2 A sin A}{2 cos 2 A sin 2 A}$

$(sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2}), cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}))$

$⟹ - tan 2 A$

$\frac{cos 2 A - cos 4 A}{sin 4 A - sin 2 A}$

$⟹ \frac{2 sin 3 A sin A}{2 cos 3 A sin 2 A}$

$(sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2}), cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}))$

$⟹ tan 3 A$

$\frac{cos 3 A - cos A}{sin 3 A - sin A} + \frac{cos 2 A - cos 4 A}{sin 4 A - sin 2 A} = \frac{cos 2 A}{sin A cos 3 A}$

$⟹ tan 3 A - tan 2 A$

$⟹ \frac{sin 3 A}{cos 3 A} - \frac{sin 2 A}{cos 2 A}$

$⟹ \frac{sin 3 A cos 2 A - cos 3 A sin 2 A}{cos 3 A cos 2 A}$

$⟹ \frac{sin (3 A - 2 A)}{cos 3 A cos 2 A}$

$(sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2}))$

$⟹ \frac{sin A}{cos 3 A cos 2 A}$

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Q. Prove that:
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(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
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sin 3A
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(v) cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −

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\frac{{cos}^{3} A + {sin}^{3} A}{cos A + sin A} + \frac{{cos}^{3} A - {sin}^{3} A}{cos A - sin A}

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