1. Right triangle has legs with lengths \(\sqrt7\) on the bottom and 1 on the right side. Use this information to find:
\(\begin{align*} &\sqrt{8}\cdot \cos\left(\arctan\left(\sqrt{7}\right)\right), \\ &\sqrt{8} \cdot \cos\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \\ &\sqrt{8} \cdot \sin\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \end{align*}\)in the same order.
2. Calculate \(\arcsin(1), \arccos\left(\frac{1}{2} \right), \arctan(1), \arcsin\left(-\frac{1}{2}\right)\) in radians and enter them below.
3. A 3-4-5 right triangle has length 3 on the bottom side and 4 on the right side. Use this information to find:
\(\sin\left(\arctan\left(\frac{3}{4}\right)\right), \cos\left(\arcsin\left(\frac{3}{5}\right)\right), \cot\left(\arctan\left(\frac{4}{3}\right)\right), \sin\left(\arcsin\left(\frac{4}{5}\right)\right)\)in the same order.
Thank you for your help! I'm having a bit of trouble with calculus.
+128406 1. √8 * cos [ arctan (√7 ]
This says that we first need to find the cosine of an angle whose tangent = y / x = √7/1
So we have x / √[y^2 + x^2 ] = 1 / √[ (√7)^2 + 1^2] = 1 / √ [7 + 1] = 1 / √8
So...
√8 * 1 / √8 = 1
√8 * cos [ arccos (1/ √8) ]
Note that cos [ arccos ( 1/√8) ] = 1/√8
So, again.... √8 * 1 / √8 = 1
√8 * sin [ arcos (1/√8)].....this says that we first need to find the sin of an angle whose cosine = 1 / √8
y = √[ r^2 - x^2] = √[ 8 - 1 ] = √7
So sin [ arccos (1 / √8 ) ] = y/r = √7/√8
So
√8 * √7/√8 = √7
+128406 2.
arcsin 1 = we are asking.....where is the sine = 1 ???......this is at pi/2 rads
arccos 1/2 = where is the cosine = 1/2 ???.....this is at pi/3 rads
arctan 1 = where is the tangent = 1 ???......this is at pi/4 rads
arcsin -1/2 = where is the sine = -1/2 ???.....this is at - pi/6 rads
P.S. - you should memorize the values of the sine, cosine and tangent at 0, pi/6, pi/4, pi/3 and pi/2 rads
+36915 Here is a sample of how to solve #3 see the image....think you can now do the others?
+128406 3.
Note that r = √[3^2 + 4^2] = √25 = 5
sin [ arctan (3/4) ] = we are looking for the sine of an angle whose tangent = 3/4 = y r = 3 / 5
cos [arcsin (3/5)] = we are looking for the cosine whose sine = 3/5 = [ 5^2 - 3^2] / 5 = x/r = √16/5 = 4/5
cot [ arctan ( 4/3) ] = we are looking for the tangent of an angle whose cot = 4/3 = 3/4
sin [arcsin (4/5) ] = 4/5
