If cos(a)=-12/13 and cot(b)=24/7,90<a<180 and 180<b<270 then the quadrant in which a+b lies?
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Expert's answer
2015-06-01T09:54:03-0400
Answer on Question #50495 – Math – Trigonometry
If cos(a)=−12/13 and cot(b)=24/7, 90∘<a<180∘ and 180∘<b<270∘ then the quadrant in which a+b lies?
Solution
Since cot(b)=sin(b)cos(b), then 1+cot2(b)=sin2(b)1=1+72242=49625, so sin2(b)=62549. Due to 180∘<b<270∘ (so, sin(b)<0 and cos(b)<0), we obtain sin(b)=−257 and cos(b)=−1−sin2(b)=−1−62549=−625576=−2524.
Due to 90∘<a<180∘ (so, sin(a)>0) we obtain sin(a)=1−cos2(a)=1−169144=16925=135.
Since sin(a+b)=sin(a)cos(b)+cos(a)sin(b)=135⋅25−24+13−12⋅25−7=25⋅13−120+84=25⋅13−36<0 and cos(a+b)=cos(a)cos(b)−sin(a)sin(b)=13−12⋅25−24−135⋅25−7=25⋅13288+35=25⋅13323>0 then we obtain that 270∘<a+b<360∘ because there exists only one quadrant where x=a+b, cosx>0 and sinx<0 simultaneously.
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