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2 years ago

If the sine of angle a is 15/17 what is the cosine of angle a

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

5 0

Answer:

sin A = opp/hyp = 15/17

Right triangle

Therefor

a² + b² = c²

adj² + opp² = hyp²

adj² + 15² = 17²

adj² = 17² - 15²

adj² = 289 - 225

adj² = 64

adj = 8

Cos = adj/hyp

cos A = 8/17 <–––––

Step-by-step explanation:

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3 years ago

Determine the intercepts of the line. 7x-5=4y-67x−5=4y−67, x, minus, 5, equals, 4, y, minus, 6 xxx-intercept: \Big((left parenth

sesenic [268]

Given:

The equation of line is

$7x-5=4y-6$

To find:

The x and y-intercept of the given line.

Solution:

We have,

$7x-5=4y-6$

Putting x=0, we get

$7(0)-5=4y-6$

Add 6 on both sides.

$0-5+6=4y$

$1=4y$

Divide both sides by 4.

$\dfrac{1}{4}=y$

So, the y-intercept is $\left(0,\dfrac{1}{4}\right)$ .

Putting y=0 in given equation, we get

$7x-5=4(0)-6$

Add 5 on both sides.

$7x=-6+5$

$7x=-1$

Divide both sides by 7.

$x=\dfrac{-1}{7}$

So, the x-intercept is $\left(-\dfrac{1}{7},0\right)$ .

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3 years ago

Read 2 more answers

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

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3 years ago