Questions on Algebra: Trigonometry answered by real tutors!

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I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
I doubled the 15 degree angle and created a 30-60-90 triangle so I know the new hypotenuse is twice the new short side, and that the common long side is root three times the short side. I also have the Pythagorean relations for the two right triangles, so I have a system of equations, but my system resolves to an identity. I need to introduce some new information.
Thanks, John

The easiest method is to use the difference of angles formula.

sin (A – B) = Sin A Cos B – Cos A Sin B

sin (60 - 45) = Sin 60 Cos 45 – Cos 60 Sin 45

sin = Sin 60 Cos 45 – Cos 60 Sin 45

Length of hypotenuse = , or (in simplest radical form)

You can do the check!!

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com

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I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
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c is the hypotenuse
sin(15) = 2/c
c = 2/sin(15)
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Use the Half-Angle formula to get sin(15)

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Remember SOHCAHTOA
Sin15=2/hypotenuse



this is ~ 7.727

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state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2
***
Equation for cos function: y=Acos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B
For given equation: y=4cos(3x+3π/2)+2
A=4
B=3
period=2π/B=2π/3
1/4 period=2π/12=π/6
C=3π/2
phase shift=C/B=(3π/2)/3=π/2 (shift to the left)=3π/6
curve is bumped up 2 units
..
Graphing function:
On an x-y coordinate system with the x-axis scaled in radians, start with coordinates of the basic cos functon with period=2π/3,amplitude =1, phase shift=0, vertical shift=0, for one period.
coordinates of basic cos function:
(0,1), (π/6,0), (π/3,-1), (π/2,0), (2π/3,1), (5π/6,0)
..
shift curve π/2 to the left:
(-π/2,1), (-π/3,0), (-π/6,-1), (0,0), (π/6,1), (π/3,0)
..
change amplitude to 4:
(-π/2,4), (-π/3,0), (-π/6,-4), (0,0), (π/6,4), (π/3,0)
..
vertical shift up 2 units (final configuration)
(-π/2,6), (-π/3,2), (-π/6,-2), (0,2), (π/6,6), (π/3,2)
set x=0
y-intercept: 4cos(3x+3π/2)+2=4cos(3π/2)+2=-4+2=2

..
It is easy to make a mistake here. Please check my calculations.

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Find the sec ɵ and tan ɵ if sin ɵ =1/5 and ɵ is in the 1st quadrant
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cos = sqrt(1 - sin^2) = sqrt(24/25) = sqrt(24)/5
sec = 1/cos = 1/sqrt(24)/5 = 5/sqrt(24)
sec = 5*sqrt(24)/24
sec = 5*sqrt(6)/12
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tan = sin/cos = sin*sec
tan = sqrt(6)/12

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The Cos of 60 degrees gives you .5, as would the Cos of -60 degrees which is the same as 300 degrees.

tan3α·cotα;

It's easy if you're taking calculus, and can use
L'Hopital's rule, but not if you aren't.  Tell me 
in the thank-you note form if you are taking 
calculus or just taking trigonometry.  Then I'll
help you with it.  Is this familiar?



Numerator and denominator both approach 0, so we
can use L'hopital's rule:



Using reciplocal identity

 = 3

That's the start of the way to do it by calculus.
I'll finish it if you tell me whether you're 
taking calculus or just trig.

Edwin