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Ierofanga [76]
3 years ago
6

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1​

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

See below.

Step-by-step explanation:

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1​

The 2 halves of this expression are the identities for [sin(A + B)]^2 and

[(cos (A + B)] ^2 respectively , therefore:

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2 =  [sin(A + B)]^2 +

[(cos (A + B)] ^2

Using the identity sin^2Ф + cos^2Ф = 1  we see that if we put Ф = (A + B) we have

 [sin(A + B)]^2 + [(cos (A + B)] ^2 = 1  so the identity

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2 = 1  must be true also.

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Zigmanuir [339]

Answer:

\frac{1}{2} (a + 2b)(a - b)

Step-by-step explanation:

Assuming you require the expression to be factored

Given

\frac{1}{2} a² + \frac{1}{2} ab - b² ← factor out \frac{1}{2} from each term

= \frac{1}{2} (a² + ab - 2b²) ← factor the quadratic

Consider the factors of the coefficient of the b² term(- 2) which sum to give the coefficient of the ab- term (+ 1)

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2 × - 1 = - 2 and 2 - 1 = + 1, thus

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\frac{1}{2} a² + \frac{1}{2} ab - b² = \frac{1}{2}(a + 2b)(a - b)

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2 years ago
Find the area of the figure
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9514 1404 393

Answer:

  24.885 in²

Step-by-step explanation:

Use the formula for the area of a triangle.

  A = 1/2bh . . . . . . . where b is the base length and h is the height perpendicular to the base

  A = 1/2(7.9 in)(6.3 in) = 24.885 in²

_____

<em>Additional comment</em>

The given side lengths cannot form a triangle, as the side shown as 14.7 is too long for the ends of the other segments to connect to. The attachment shows that side should be 11.97 in.

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