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3 years ago

Q.23 Prove that: (sin4A - cos4A + 1) = 2sin2A

Mathematics

1 answer:

FrozenT [24]3 years ago

4 0

Answer:

Step-by-step explanation:

I think you have the question incomplete, and that this is the complete question

sin^4a + cos^4a = 1 - 2sin^2a.cos^2a

To do this, we can start my mirroring the equation.

x² + y² = (x + y)² - 2xy,

This helps us break down the power from 4 to 2, so that we have

(sin²a)² + (cos²a)² = (sin²a + cos²a) ² - 2(sin²a) (cos²a)

Recall from identity that

Sin²Φ + cos²Φ = 1, so therefore

(sin²a)² + (cos²a)² = 1² - 2(sin²a) (cos²a)

On expanding the power and the brackets, we find that we have the equation proved.

sin^4a + cos^4a = 1 - 2sin^2a.cos^2a

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Step-by-step explanation:

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$As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel$

$Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm$

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3 years ago

Sin (3x+15) = Cos (7x-2)<br><br> Find value of x

11Alexandr11 [23.1K]

$\bf \textit{Cofunction Identities}
\\ \quad \\
\boxed{sin\left(\frac{\pi}{2}-{{ \theta}}\right)=cos({{ \theta}})}\qquad 
cos\left(\frac{\pi}{2}-{{ \theta}}\right)=sin({{ \theta}})
\\ \quad \\ \quad \\
tan\left(\frac{\pi}{2}-{{ \theta}}\right)=cot({{ \theta}})\qquad 
cot\left(\frac{\pi}{2}-{{ \theta}}\right)=tan({{ \theta}})
\\ \quad \\ \quad \\
sec\left(\frac{\pi}{2}-{{ \theta}}\right)=csc({{ \theta}})\qquad 
csc\left(\frac{\pi}{2}-{{ \theta}}\right)=sec({{ \theta}})\\\\
-------------------------------\\\\$

$\bf \begin{array}{lcllclll}
sin(&90^o-\theta)&=&cos(&\theta)\\
&\uparrow &&&\uparrow \\
&3x+15&&&7x-2\\
&\downarrow \\
&90-(7x-2)
\end{array}
\\\\\\
90-(7x-2)=3x+15\implies 90-7x+2=3x+15
\\\\\\
92-15=10x\implies 77=10x\implies \cfrac{77}{10}=x$

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