Question

calculati:a)2sin30-3tg30+4cos60.,b) sin45-ctg30+2cos60-cos45.,c)(1+tg²60°)•cos45°-sin45°•sin²60°"

Answer 1

a)$2sin30-3tg30+4cos60= \\ =2* \frac{1}{2}-3* \frac{1}{ \sqrt{3} }+4* \frac{1}{2}=1- \frac{3}{ \sqrt{3} }+2=3- \frac{3 \sqrt{3} }{3}=3- \sqrt{3}$

b) $sin45-ctg30+2cos60-cos45=\frac{ \sqrt{2} }{2}- \sqrt{3}+2* \frac{1}{2}- \frac{ \sqrt{2} }{2}= - \sqrt{3}+1$

c) $(1+ tg^{2} 60)cos45-sin45*sin^260= \\ =[1+( \sqrt{3})^2] \frac{ \sqrt{2} }{2}- \frac{ \sqrt{2} }{2}*( \frac{ \sqrt{3} }{2})^2= \\ =(1+3) \frac{ \sqrt{2} }{2}- \frac{ \sqrt{2} }{2}* \frac{3}{4}= \frac{4 \sqrt{2} }{2}- \frac{3 \sqrt{2} }{8}= \frac{16 \sqrt{2}-3 \sqrt{2} }{8}= \frac{13 \sqrt{2} }{8}$