Prove that sinA - sin3A - sin5A - sin7AcosA - cos 3A - cos 5A - cos 7A - tan 4A

L.H.S #10; #10; =sin A+sin 3A+sin 5A+sin 7Acos A+cos #10;3A+5A+cos 7A #10; #10; #160; #10; #10;We know that #10; #10; sin C+sin D=2 ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Prove that s...

Question

Prove that
$\frac{sinA + s i n 3 A + s i n 5 A + s i n 7 A}{c o s A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A$

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Solution

L.H.S

$= \frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + 5 A + cos 7 A}$

We know that

$sin C + sin D = 2 sin \frac{C + D}{2} \cdot cos \frac{C - D}{2}$

Therefore,

$= \frac{2 sin (\frac{A + 3 A}{2}) \cdot cos (\frac{A - 3 A}{2}) + 2 sin (\frac{5 A + 7 A}{2}) \cdot cos (\frac{5 A - 7 A}{2})}{2 cos (\frac{A + 3 A}{2}) \cdot cos (\frac{A - 3 A}{2}) + 2 cos (\frac{5 A + 7 A}{2}) \cdot cos (\frac{5 A - 7 A}{2})}$

$= \frac{2 sin (2 A) \cdot cos (- A) + 2 sin (6 A) \cdot cos (- A)}{2 cos (2 A) \cdot cos (- A) + 2 cos (6 A) \cdot cos (- A)}$

$= \frac{2 sin (2 A) \cdot cos (A) + 2 sin (6 A) \cdot cos (A)}{2 cos (2 A) \cdot cos (A) + 2 cos (6 A) \cdot cos (A)} [cos (- θ) = cos θ]$

$= \frac{sin (2 A) + sin (6 A)}{cos (2 A) + cos (6 A)}$

$= \frac{sin (2 A) + sin (6 A)}{cos (2 A) + cos (6 A)}$

$= \frac{2 sin (\frac{2 A + 6 A}{2}) \cdot cos (\frac{2 A - 6 A}{2})}{2 cos (\frac{2 A + 6 A}{2}) \cdot cos (\frac{2 A - 6 A}{2})}$

$= \frac{sin (4 A)}{cos (4 A)}$

$= tan 4 A$

Hence, proved.

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Similar questions

\frac{c o s A + c o s 3 A + c o s 5 A + c o s 7 A}{s i n A + s i n 3 A + s i n 5 A + s i n 7 A} =

Q. (i)

\frac{cos 3 θ + 2 cos 5 θ + cos 7 θ}{cos θ + 2 cos 3 θ + cos 5 θ} = cos 2 θ - sin 2 θ tan 3 θ

(ii)

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}

is equal to -

Q. Solve the problem:-

\frac{sin A - sin 3 A + sin 5 A - sin 7 A}{cos A - cos 3 A - cos 5 A + cos 7 A} = 2 cot A

8. Prove that :
(i) $\frac{s i n A + s i n 3 A + s i n 5 A}{c o s A + c o s 3 A + c o s 5 A} = t a n 3 A$
(ii) $(i i) \frac{c o s 3 A + 2 c o s 5 A + c o s 7 A}{c o s A + 2 c o s 3 A + c o s 5 A} = \frac{c o s 5 A}{c o s 3 A}$
(iii) $\frac{c o s 4 A + c o s 3 A + c o s 2 A}{c o s 4 A + s i n 3 A + s i n 2 A} = c o t 3 A$
(iv) $\frac{s i n 3 A + s i n 5 A + s i n 7 A + s i n 9 A}{c o s 3 A + c o s 5 A + c o s 7 A + c o s 9 A} = t a n 6 A$
(v) $\frac{s i n 5 A - s i n 7 A + s i n 8 A - s i n 4 A}{c o s 4 A + c o s 7 A + c o s 7 A - c o s 5 A - c o s 8 A} = c o t 6 A$
(vi) $\frac{s i n 5 A + c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A} = t a n A$
(vii) $\frac{s i n 11 A + s i n A + s i n 7 A + s i n 3 A}{c o s 11 A s i n A + c o s 7 A s i n 3 A} = t a n 8 A$
(viii) $\frac{s i n 3 A c o s 4 A - s i n A c o s 2 A}{s i n 4 A s i n A + c o s 6 A c o s A} = t a n 2 A$
(ix) $\frac{s i n A s i n 2 A + s i n 3 A s i n 6 A}{s i n A c o s 2 A + c o s 3 A c o s 6 A} = t a n 5 A$
(x) $\frac{s i n A + 2 s i n 3 A + s i n 5 A}{s i n 3 A + 2 s i n 5 A + s i n 7 A} = \frac{s i n 3 A}{s i n 5 A}$
(xi) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

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Prove thatsinA+sin3A+sin5A+sin7AcosA+cos3A+cos5A+cos7A=tan4A

Prove that
$\frac{sinA + s i n 3 A + s i n 5 A + s i n 7 A}{c o s A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A$