Răspuns:
1.a
2cos30-tg60+√2cos45°=
2*√3/2-√3+√2*√2/2=
√3-√3+2/2=0+1=1
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b) sin90°+tg 45°/sin60°=
1+1/(√3/2)=1+2/√3=amplifici cu √3=
1+2√3/3
2-cos15=(√6+√2)/4
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Aplici formula
cosx=cos²/2-sim²x/2
cos30=cos²15°-sin²15=√3/2
cos²15°-(1-cos²15)=√3/2
cos²15°-1+cos²15°=√3/2
2cos²15=√3/2+1
2cos²15=(√3+2)/2
cos²15=(√3+2)/4
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3)sin²30°+cos²60°=
(1/2)²+(√3/2)²=
1/4+3/4=1
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4)Aplici teorema cosinusului
BC²=AB²+AC²-2AB*AC*cosA
2²=3²+4²-2*3*4cosA
4=9+16-24cos<A
4=25-24cos<A
4-25= -24cos<A
-21=-24 cos<A
cos<A= -21/24
cos <A= -7/8
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5.Aplici teorema sinudsului
BC=a R=raza
a/sinA=2R
7√3/sin60=2R
7√3/(√3/2)=2R
14√3/√3=2R
2R=14
R=7
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6.sinx =5/13
Aplici formula
sin²x+cos²x=1
(5/13)²+cos²x=1
25/169+cos²x=1
cos²x=1-25/169
cos²x=144/169
cosx=√144/169=±12/13
Deoarece x∈(π/2,π) se ia valoarea negativa
cos x= -12/13
Explicație pas cu pas:
