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 May 15th, 2018, 01:13 PM #1 Senior Member   Joined: Jan 2017 From: US Posts: 111 Thanks: 6 Solve for the value of sin Could anyone point me in the right direction of solving for the value of sin 150 degrees without using a calculator? I have to come up with an exact value, not a rounded decimal. Any help would be greatly appreciated!
May 15th, 2018, 02:04 PM   #2
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reference angle is 30 degrees in quad II, forming a 30-60-90 triangle.

what do you recall about the side ratios for such a triangle?
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 May 16th, 2018, 02:41 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,248 Thanks: 887 150= 90+ 60. Now, think about an equilateral triangle. It has all thee sides the same length, s, say, and all three angle with measure 60 degrees. If you drop a perpendicular from one vertex to the opposite side, it divides the equilateral triangle into to right angles with angles 60 and 30 degrees. The hypotenuse has length s, one side has length s/2 and you can calculate the third side using the Pythagorean theorem.
 May 16th, 2018, 05:43 AM #4 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 606 Thanks: 82 I used skeeter's diagram and my memory about side ratios in a 30-60-90 triangle to calculate cos 30, which equals sin 150.
May 16th, 2018, 12:54 PM   #5
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Quote:
 Originally Posted by EvanJ I used skeeter's diagram and my memory about side ratios in a 30-60-90 triangle to calculate cos 30, which equals sin 150.
sin(150)=sin(30), not cos(30). Your approach still works.

 May 17th, 2018, 12:00 PM #6 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 606 Thanks: 82 I figured out my mistake. I thought sin x = cos (180-x) when it is sin x = cos (90-x).
 June 3rd, 2018, 01:05 PM #7 Newbie   Joined: Dec 2017 From: Spain Posts: 16 Thanks: 1 For common angles, like 0º, 30º, 45º, 60º and 90º, there are special sine-values. Since sin(150)=sin(30)=0.5 Remember: sin(0)= 0.5*[sqrt(0)] sin(30)=0.5*[sqrt(1)]=0.5 sin(45)=0.5*[sqrt(2)] sin(60)=0.5*[sqrt(3)] sin(90)=0.5*[sqrt(4)]=1 Look in this page: Sines and Cosines of Common Angles -- The Brain Dump
 June 4th, 2018, 09:09 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,190 Thanks: 1649 sin(15$^\circ\!$) = 0.25(√6 - √2) sin(18$^\circ\!$) = 0.25(√5 - 1) sin(54$^\circ\!$) = 0.25(√5 + 1) sin(75$^\circ\!$) = 0.25(√6 + √2)

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