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3 years ago

1)calculate please it 2) prove it please

Mathematics

1 answer:

Dvinal [7]3 years ago

4 0

1. 2sin²(-2490) + tan(1410)
2(0.25) - 0.6
0.5 + 0.6
1.1

2. $\frac{2sin(\alpha)cos(\beta) - sin(\alpha - \beta)}{cos(\alpha - \beta) - 2sin(\alpha)sin(b)} = tan(\alpha + \beta)$
   $\frac{2sin(\alpha)cos(\beta) - sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)}{cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)} = tan(\alpha + \beta)$
   $\frac{sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)}{cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)} = tan(\alpha + \beta)$
   $\frac{sin(\alpha + \beta)}{cos(\alpha + \beta)} = tan(\alpha + \beta)$
   $tan(\alpha + \beta) = tan(\alpha + \beta)$

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Answer:

$y=2e^{sin(x)}$

Step-by-step explanation:

Given equation can be re written as

$\frac{\mathrm{d} y}{\mathrm{d} x}-ycos(x)=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=ycos(x)\\\\=> \frac{dy}{y}=cox(x)dx\\\\Integrating \\ \int \frac{dy}{y}=\int cos(x)dx \\\\ln(y)=sin(x)+c$ ............(i)

Now it is given that y(π/2) = 2e

Applying value in (i) we get

ln(2e) = sin(π/2) + c

=> ln(2) + ln(e) = 1+c

=> ln(2) + 1 = 1 + c

=> c = ln(2)

Thus equation (i) becomes

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$y= 2e^{sinx}$

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3 years ago