Help 3uniy intergrand (1 Viewer)

petergriffin198 · Oct 23, 2015

Help both

rand_althor · Oct 23, 2015

$$Draw a right-angled triangle in the second quadrant with $\sin{x}=k$. From this, you can find that $\cos{x}=-\sqrt{1-k^2}$. Now, $\tan{\left (\frac{\pi}{2}+x \right )}=-\cot{x}=\frac{-\cos{x}}{\sin{x}}=\frac{\sqrt{1-k^2}}{k}$.$

Edit: made a mistake.

godofindolence · Oct 23, 2015

rand_althor said:
$$Draw a right-angled triangle in the second quadrant with $\sin{x}=k$. From this, you can find that $\cos{x}=-\sqrt{1-k^2}$. Now, $\tan{\left (\frac{\pi}{2}+x \right )}=-\cot{x}=\frac{-\cos{x}}{\sin{x}}=\frac{-\sqrt{1-k^2}}{k}$.$

Positive as in third quadrant?

integral95 · Oct 23, 2015

For the first question,

construct a right angle triangle according sinx = k

now

$tan(\frac{\pi}{2} + x ) = \frac{sin(\frac{\pi}{2} + x )}{cos(\frac{\pi}{2} + x )} = \\ \\ \\ \frac{-cosx}{sinx} = -\frac{-\sqrt{1-k^2}}{k} = \\ \\ \\ \frac{\sqrt{1-k^2}}{k}$

InteGrand · Oct 23, 2015

petergriffin198 said:
View attachment 32591

Help both

$For the second question:$

$$g(x)+a-x=0$. Let $f(x) := g(x) + a-x$, so $f^\prime (x) = g^\prime (x) -1$ and the equation to solve is $f(x) = 0$.$

$The first guess is $x_0 = a$.$

$The next approximation is $x_1 = x_0 - \frac{f(x_0)}{f^\prime (x_0)}$$

$\Rightarrow x_1 = a- \frac{f(a)}{f^\prime (a)}$

$=a-\frac{g(a)+a-a}{g^\prime (a) - 1}$

$=a-\frac{g(a)}{g^\prime (a) - 1}$

$$=a+\frac{g(a)}{1- g^\prime (a) }$.$

$Hence the answer is (A).$

Mikes · Oct 23, 2015

integral95 said:
For the first question,

construct a right angle triangle according sinx = k

now

$tan(\frac{\pi}{2} + x ) = \frac{sin(\frac{\pi}{2} + x )}{cos(\frac{\pi}{2} + x )} = \\ \\ \\ \frac{-cosx}{sinx} = -\frac{-\sqrt{1-k^2}}{k} = \\ \\ \\ \frac{\sqrt{1-k^2}}{k}$

How did you go from sin(pi/2+x) to -cosx?

InteGrand · Oct 23, 2015

Mikes said:
How did you go from sin(pi/2+x) to -cosx?

$This is a trigonometric identity that can be proved in many ways, e.g. the graph of $y=\sin \left(\frac{\pi}{2}+x \right)$ is that of $y=\sin x$ shifted left by $\frac{\pi}{2}$ units, which we observe is just the graph of $y=\cos x$, which means that $\cos x = \sin \left(\frac{\pi}{2}+x \right)$ for all real $x$.$

InteGrand · Oct 23, 2015

Mikes said:
How did you go from sin(pi/2+x) to -cosx?

$If you want an algebraic proof, here is one:$

$$\sin \left(\frac{\pi}{2}+x \right) = \cos \left(\frac{\pi}{2}- \left(\frac{\pi}{2}+x \right) \right)$ (complementary angles trig. identity)$

$$=\cos \left(-x \right)$$

$$=\cos x$ (cosine function is even).$

$(The identity is $\sin \left(\frac{\pi}{2}+x \right)=\cos x$, not $-\cos x$.)$

Mikes · Oct 23, 2015

InteGrand said:
$If you want an algebraic proof, here is one:$

$$\sin \left(\frac{\pi}{2}+x \right) = \cos \left(\frac{\pi}{2}- \left(\frac{\pi}{2}+x \right) \right)$ (complementary angles trig. identity)$

$$=\cos \left(-x \right)$$

$$=\cos x$ (cosine function is even).$

$(The identity is $\sin \left(\frac{\pi}{2}+x \right)=\cos x$, not $-\cos x$.)$

Thanks for clarifying. I was confused because OP had it as sin(pi/2+x)=-cosx

InteGrand · Oct 23, 2015

Mikes said:
Thanks for clarifying. I was confused because OP had it as sin(pi/2+x)=-cosx

$No he didn't actually (assuming you're referring to integral95's answer). His denominator from the previous line becomes $-\sin x$ (another identity), so he just moved the minus sign of this to the numerator in his next line, which is valid to do (though I can understand why you could be confused by that).$

petergriffin198 · Oct 23, 2015

Why is it positive sinx not negative?

InteGrand · Oct 23, 2015

petergriffin198 said:
Why is it positive sinx not negative?

It's not, he just moved the minus sign to the numerator.

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