Answer:
Correct option is
B
0
D
−1
sinx+sin2x+sin3x
=sin(2x−x)+sin2x+sin(2x+x)
=2sin2xcosx+sin2x [ by using sin(A+B)=sinAcosB+sinBcosA and sin(A−B)=sinAcosB−sinBcosA ]
=sin2x(2cosx+1)........(i)
cosx+cos2x+cos3x
=cos(2x−x)+cos2x+cos(2x+x)
=2cos2xcosx+cos2x [By using cos(a−b)=cosa⋅cosb+sina⋅sinb and cos(a+b)=cosa⋅cosb−sina⋅sinb]
=cos2x(2cosx+1).....(ii)
∴(sinx+sin2x+sin3x)
2
+(cosx+cos2x+cos3x)
2
=1
sin
2
2x(2cosx+1)
2
+cos
2
2x(2cosx+1)
2
=1.......[From(i)(ii)]
⇒(2cosx+1)
2
=1
⇒2cosx+1=±1
∴cosx=0or−1
Answer:5
Step-by-step explanation:
Add 6 and 4
2*10-3*5
Multiply 2 by 10
20-3*5
Multiply -3 by 5
20-15
Subtract 15 from 20
5
Answer:
The one on the left
Step-by-step explanation:
i thought about it and its the dot on the left
Answer:
X = 16
Step-by-step explanation:
Notice that we are in the presence of two similar triangles, one larger than the other. The larger triangle is AED, and the smaller one is ACB.
Since there are similar triangles, we can write a proportionality between their known sides:
notice that AD = x + 6
AB = x
ED = 11
CB = 8
So now we can re-write the proportion and solve it for the unknown "x":
