1) Use half-angle identity for sine:
s i n a 2 = ± 1 − c o s a 2 sin \frac{a}{2}=±\sqrt \frac{1-cos a}{2} s in 2 a = ± 2 1 − cos a Find c o s a : cos a: cos a :
c o s a = ± 1 − s i n 2 a cos a=±\sqrt{1-sin^2 a} cos a = ± 1 − s i n 2 a c o s a = ± 1 − ( 4 5 ) 2 = ± 1 − 16 25 = ± 9 25 = ± 3 5 cos a=±\sqrt{1-(\frac{4}{5})^2}=±\sqrt{1-\frac{16}{25}}=±\sqrt{\frac{9}{25}}=±\frac{3}{5} cos a = ± 1 − ( 5 4 ) 2 = ± 1 − 25 16 = ± 25 9 = ± 5 3 Since a a a is in the quadrant I, c o s a > 0 , c o s a = 3 5 cos a>0, cos a=\frac{3}{5} cos a > 0 , cos a = 5 3
Hence,
s i n a 2 = ± 1 − 3 5 2 = ± 2 5 2 = ± 1 5 sin \frac{a}{2}=±\sqrt{\frac{1-\frac{3}{5}}{2}}=±\sqrt{\frac{\frac{2}{5}}{2}}=±\sqrt{\frac{1}{5}} s in 2 a = ± 2 1 − 5 3 = ± 2 5 2 = ± 5 1 Since a a a is in the quadrant I, s i n a 2 > 0 , s i n a 2 = 1 5 sin \frac{a}{2}>0, sin \frac{a}{2}=\sqrt{\frac{1}{5}} s in 2 a > 0 , s in 2 a = 5 1
2) Use half-angle identity for cosine:
c o s a 2 = ± 1 + c o s a 2 cos \frac{a}{2}=±\sqrt \frac{1+cos a}{2} cos 2 a = ± 2 1 + cos a c o s a 2 = ± 1 + 3 5 2 = ± 8 5 2 = ± 4 5 cos \frac{a}{2}=±\sqrt{\frac{1+\frac{3}{5}}{2}}=±\sqrt{\frac{\frac{8}{5}}{2}}=±\sqrt{\frac{4}{5}} cos 2 a = ± 2 1 + 5 3 = ± 2 5 8 = ± 5 4
Since a a a is in the quadrant I, c o s a 2 > 0 , c o s a 2 = 4 5 cos \frac{a}{2}>0, cos \frac{a}{2}=\sqrt{\frac{4}{5}} cos 2 a > 0 , cos 2 a = 5 4
3) Half-angle for tangent and cotangent:
t a n a 2 = s i n a 2 : c o s a 2 = 1 5 : 4 5 = 1 4 = 1 2 tan \frac{a}{2}=sin \frac{a}{2}:cos \frac{a}{2}=\sqrt{\frac{1}{5}}:\sqrt{\frac{4}{5}}=\sqrt{\frac{1}{4}}=\frac{1}{2} t an 2 a = s in 2 a : cos 2 a = 5 1 : 5 4 = 4 1 = 2 1 c o t a 2 = c o s a 2 : s i n a 2 = 4 5 : 1 5 = 4 1 = 2 cot \frac{a}{2}=cos \frac{a}{2}:sin \frac{a}{2}=\sqrt{\frac{4}{5}}:\sqrt{\frac{1}{5}}=\sqrt{\frac{4}{1}}=2 co t 2 a = cos 2 a : s in 2 a = 5 4 : 5 1 = 1 4 = 2
4) Use double-angle identity for sine:
s i n 2 a = 2 s i n a c o s a sin 2a=2sinacos a s in 2 a = 2 s ina cos a s i n 2 a = 2 ⋅ 4 5 ⋅ 3 5 = 24 25 sin 2a=2\cdot \frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25} s in 2 a = 2 ⋅ 5 4 ⋅ 5 3 = 25 24 5) Use double-angle identity for cosine:
c o s 2 a = 1 − 2 s i n 2 a = 1 − 2 ⋅ ( 4 5 ) 2 = 1 − 32 25 = − 7 25 cos 2a=1-2sin^2a=1-2\cdot({\frac{4}{5}})^2=1-\frac{32}{25}=-\frac{7}{25} cos 2 a = 1 − 2 s i n 2 a = 1 − 2 ⋅ ( 5 4 ) 2 = 1 − 25 32 = − 25 7 6) Double-angle for tangent and cotangent:
t a n 2 a = s i n 2 a : c o s 2 a = 24 25 : ( − 7 25 ) = − 24 7 tan2a=sin2a:cos 2a=\frac{24}{25}:(-\frac{7}{25})=-\frac{24}{7} t an 2 a = s in 2 a : cos 2 a = 25 24 : ( − 25 7 ) = − 7 24
c o t 2 a = c o s 2 a : s i n 2 a = − 7 25 : 24 25 = − 7 24 cot2a=cos2a:sin 2a=-\frac{7}{25}:\frac{24}{25}=-\frac{7}{24} co t 2 a = cos 2 a : s in 2 a = − 25 7 : 25 24 = − 24 7
Answer: s i n a 2 = 1 5 , c o s a 2 = 4 5 , t a n a 2 = 1 2 , c o t a 2 = 2 , sin \frac{a}{2}=\sqrt{\frac{1}{5}}, cos \frac{a}{2}=\sqrt{\frac{4}{5}}, tan \frac{a}{2}=\frac{1}{2}, cot\frac{a}{2}=2, s in 2 a = 5 1 , cos 2 a = 5 4 , t an 2 a = 2 1 , co t 2 a = 2 ,
s i n 2 a = 24 25 , c o s 2 a = − 7 25 , t a n 2 a = − 24 7 , c o t 2 a = − 7 24 sin 2a=\frac{24}{25}, cos 2a=-\frac{7}{25}, tan 2a=-\frac{24}{7}, cot 2a=-\frac{7}{24} s in 2 a = 25 24 , cos 2 a = − 25 7 , t an 2 a = − 7 24 , co t 2 a = − 24 7
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