Question

cos 6 theta +6 cos 3 theta +15 cos 2 theta+10 /cos 5 theta +5 cos 3 theta +10
1) sin theta 2) cos theta 3) 2 sin theta 4) 2 cos theta

Answer 1

Answer:

Option(4) 2cosβ

Step-by-step explanation:

I think your question is $\frac{cos6\beta+6cos4\beta+15cos2\beta+10}{cos5\beta+5cos3\beta+10cos\beta}$

= $\frac{cos3(2\beta)+6cos2(2\beta)+15cos2\beta+10}{(cos5\beta+cos3\beta)+4(cos3\beta+cos\beta)+6cos\beta}$

= $\frac{4cos^{3}2\beta-3cos2\beta+6(2cos^{2}2\beta-1)+15cos2\beta+10}{2cos4\beta.cos\beta+4(2cos2\beta.cos\beta)+6cos\beta}$

= $\frac{4cos^{3}2\beta+12cos^{2}2\beta+12cos2\beta+4}{2cos\beta(cos4\beta+4cos2\beta+3)}$

= $\frac{(cos2\beta+1)(4cos^{2}2\beta+8cos2\beta+4)}{2cos\beta(2cos^{2}2\beta-1+4cos2\beta+3)}$

= $\frac{4(cos2\beta+1)(cos^{2}2\beta+2cos2\beta+1)}{2cos\beta(2cos^{2}2\beta+4cos2\beta+2)}$

= $\frac{4(cos2\beta+1)(cos2\beta+1)^{2}}{4cos\beta(cos^{2}2\beta+2cos2\beta+1)}$

= $\frac{(1+cos2\beta)^{3}}{cos\beta(1+cos2\beta)^{2}}$

= $\frac{1+cos2\beta}{cos\beta}$

= $\frac{2cos^{2}\beta}{cos\beta}$

= 2cosβ

Answer 2

Answer:

4) 2cosθ

Step-by-step explanation:

The question is cos6θ + 6cos3θ + 15cos2θ + 10/cos5θ + 5cos3θ + 10

now for the numerator:

cos6θ + 6cos3θ + 15cos2θ + 10

= cos6θ + (cos3θ + 5cos3θ) + (5cos2θ + 10cos2θ) + 10    (splitting the terms)

= 2cos5θ . cosθ + 5(2cos3θ . cosθ) + 10(2cosθ)       (taking common terms and middle term factorization)

= 2cosθ ( cos5θ + 5cos3θ + 10)    (taking common terms)

now, putting the simplified numerator on its place, we get;

2cosθ (cos5θ + 5cos3θ + 10)/ (cos5θ + 5cos3θ + 10)

(cancelling the common terms from both numerator and denominator, we get;

2cosθ; answer.