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3 years ago

(cosA+cosB)²+(sinA+sinB)²=4cos²(A-B/2)

1 answer:

3 0

Solution:
LHS = (cosA+cosB)²+(sinA+sinB)²
= (cos²A+2cosAcosB+cos²B)+
(sin²A+2sinAsinB+sin²B)
/* By algebraic identity:
(x+y)² = x²+2xy+y² */
Rearranging the terms, we get
= (cos²A+sin²A)+(cos²B+sin²B)+2(cosAcosB+sinAsinB)
= 1+1+2cos(A-B)
________________________
we know that,
i) cos²x + sin²x = 1
ii) cosxcosy+sinxsiny = sin(x-y)
________________________
= 2+2cos(A-B)
= 2[1+cos(A-B)]

=
=
=$RHS$
Therefore,
(cosA+cosB)²+(sinA+sinB)²
=
••••

diavinad8 and 60 more users found this answer helpful

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The end of the second year can be taken as the beginning of the third year.

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$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

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The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

3 years ago

(cosA+cosB)²+(sinA+sinB)²=4cos²(A-B/2)​

(cosA+cosB)²+(sinA+sinB)²=4cos²(A-B/2)