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3 years ago

For the right triangle shown, which of these equations is correct?

Mathematics

2 answers:

Shkiper50 [21]3 years ago

8 0

Answer:

3. cos(53)=sin(37)

Step-by-step explanation:

let y be the third side

cos(53)= adj/hyp

=y/16

sin(37)=opp/hyp

=y/16

therefore, sin(37)=cos(53)=y/16

klasskru [66]3 years ago

7 0

Answer:

Step-by-step explanation:

3 because 37° and 53° these sum is 90°.

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4 years ago

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2822772

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\
=\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}$

Make the following substitution:

$\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}$

and then, the integral (i) becomes

$=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}$

Integrate it by applying the power rule:

$\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\
\mathsf{=-\,\dfrac{1}{u}+C}$

Now, substitute back for u = sin x, so the result is given in terms of x:

$\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

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4 years ago