Necessary /Sufficient - Trigo

Geometry Level 4

Consider the following two statements:

P: 1 + sin ⁡ θ = a \sqrt { 1+ \sin{ \theta } } =a

Q: sin ⁡ θ 2 + cos ⁡ θ 2 = a \sin { \cfrac { \theta }{ 2 } } +\cos { \cfrac { \theta }{ 2 } } =a

Which of the following statements is true?

(A) P is necessary and sufficient for Q.

(B) P is necessary but not sufficient for Q.

(C) P is sufficient but not necessary for Q.

(D) P is neither necessary nor sufficient for Q.

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A C B D

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Soumo Mukherjee by Soumo Mukherjee , 25, India

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3 solutions

Ayush Verma
Apr 4, 2015

( 1 ) I f a = 1 + sin ⁡ θ = ( sin ⁡ θ 2 + cos ⁡ θ 2 ) 2 ⇒ a = ∣ sin ⁡ θ 2 + cos ⁡ θ 2 ∣ ( 2 ) I f a = sin ⁡ θ 2 + cos ⁡ θ 2 ⇒ a 2 = 1 + sin ⁡ θ ⇒ 1 + sin ⁡ θ = ∣ a ∣ S o w e c a n c o n c l u d e t h a t n o n e o f t h e s t a t e m e n t s a r e n e c e s s a r y o r s u f f i c i e n t f o r e a c h o t h e r \left( 1 \right) If\quad a=\sqrt { 1+\sin { \theta } } \\ \\ =\sqrt { { \left( \sin { \cfrac { \theta }{ 2 } } +\cos { \cfrac { \theta }{ 2 } } \right) }^{ 2 } } \\ \\ \Rightarrow a=\left| \sin { \cfrac { \theta }{ 2 } } +\cos { \cfrac { \theta }{ 2 } } \right| \\ \\ \\ \left( 2 \right) If\quad a=\sin { \cfrac { \theta }{ 2 } } +\cos { \cfrac { \theta }{ 2 } } \\ \\ \Rightarrow { a }^{ 2 }=1+\sin { \theta } \\ \\ \Rightarrow \sqrt { 1+\sin { \theta } } =\left| a \right| \\ \\ So\quad we\quad can\quad conclude\quad that\quad none\quad of\quad the\quad statements\quad are\quad \\ \\ necessary\quad or\quad sufficient\quad for\quad each\quad other

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Despicable Tamim
Apr 24, 2015

both of the equation is same statement .... so we can say none of them is necessary or sufficient for each other

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Soumo Mukherjee
Apr 3, 2015

Since 1 + sin ⁡ θ \sqrt { 1+ \sin{ \theta } } is never negative while sin ⁡ θ 2 + cos ⁡ θ 2 \sin { \cfrac { \theta }{ 2 } } +\cos { \cfrac { \theta }{ 2 } } may take negative values, P is neither necessary nor sufficient for Q.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I contest your point that the square root of 1+ sin ⁡ θ \sin \theta cannot be negative, if you say so you miss to add for a>0 as a premise, indeed eliminating a in both equations you get an identity.

Mariano PerezdelaCruz - 6 years, 2 months ago

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