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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,756 Thanks: 1407 
reference angle is 30 degrees in quad II, forming a 306090 triangle. what do you recall about the side ratios for such a triangle? 
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 306090 triangle to calculate cos 30, which equals sin 150.

May 16th, 2018, 12:54 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,542 Thanks: 592  
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 (180x) when it is sin x = cos (90x).

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 sinevalues. 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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