quick question on the math destroyer

[spoog74]

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so the question goes;

if csc a = - Square root of 2 find COS a in the third quadrant.


Ok great, so i figured out the value which was square root of 2 over 2, but in the third quadrant cos is negative, yet the actually value is ALREADY negative so dont you negate both negatives to give you a positive answer?

The math destroyer gives - square root of 2 / 2 as the answer, but i think it is just square root of 2/ 2 ....

Hope this makes senses..

Btw its question # 3 on practice test 9

 

[PhansterZ]

Third quadrant means cos will be negative. so negative root 2/2 is correct.

If they asked you, what value from pi -> 3pi/2 would cos = -root 2/2, you would say 5pi/4 .

so what is cos( 5pi/4)? that's -root2/2.

is that an okay explanation?

 

[nimman1126]

Since csc(a)= 1/sin you would get sin(a)= -root 2 over 2. I made a triangle in the 3rd quadrant and put that information in to complete the triangle. Then I took the cosine (Adjacent/Hypotenuse) and it comes out to -root 2 over 2

 

[goodie]

When I drew this triangle in the 3rd quadrant i got the answer to be -1/root 2 because its a 45-45-90 triangle with the rule of the sides being 1:1:root two. If I find the cos of that triangle I dont get root2/2....where am I going wrong?