find $k^2$ if $left| asin^2 theta+bsin theta cos theta+ccos^2 theta-frac{(a+c)}{2} right|=frac{k}{2}$
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If $
left|
asin^2theta+bsinthetacostheta+ccos^2theta-dfrac{(a+c)}{2}
right|=dfrac{k}{2}$, then find $k^2.$
My Attempt
begin{align}
pm k &= 2asin^2 theta+2bsin theta cos theta+2ccos^2 theta-(a+c) \
&= a[1-cos 2theta]+bsin 2theta+c[1+cos 2theta]-(a+c) \
&= aleft[ 1-frac{1-tan^2 theta}{1+tan^2 theta} right]+
bfrac{2tan theta}{1+tan^2 theta}+
cleft[ 1+frac{1-tan^2 theta}{1+tan^2 theta} right]-(a+c)\
pm k[1+tan^2 theta] &=
2atan^2 theta+2btan theta+2c-(a+c)[1+tan^2 theta] \
0 &= tan^2 theta [a-cmp k]+2btan theta+(c-amp k) \
Delta &= 4b^2-4(a-cmp k)(c-amp k) \
&= 4b^2-4[mp k+(a-c)][mp k-(a-c)] \
&= 4b^2-4[k^2-(a-c)^2] \
& ge 0 \
b^2+(a-c)^2 & ge k^2
end{align}
How do I prove that $k^2=b^2+(a-c)^2$ ?
trigonometry
edited 7 hours ago
Ng Chung Tak
13.5k31234
asked 8 hours ago
ss1729
1,613722
add a comment |
up vote
1
down vote
favorite
If $
left|
asin^2theta+bsinthetacostheta+ccos^2theta-dfrac{(a+c)}{2}
right|=dfrac{k}{2}$, then find $k^2.$
My Attempt
begin{align}
pm k &= 2asin^2 theta+2bsin theta cos theta+2ccos^2 theta-(a+c) \
&= a[1-cos 2theta]+bsin 2theta+c[1+cos 2theta]-(a+c) \
&= aleft[ 1-frac{1-tan^2 theta}{1+tan^2 theta} right]+
bfrac{2tan theta}{1+tan^2 theta}+
cleft[ 1+frac{1-tan^2 theta}{1+tan^2 theta} right]-(a+c)\
pm k[1+tan^2 theta] &=
2atan^2 theta+2btan theta+2c-(a+c)[1+tan^2 theta] \
0 &= tan^2 theta [a-cmp k]+2btan theta+(c-amp k) \
Delta &= 4b^2-4(a-cmp k)(c-amp k) \
&= 4b^2-4[mp k+(a-c)][mp k-(a-c)] \
&= 4b^2-4[k^2-(a-c)^2] \
& ge 0 \
b^2+(a-c)^2 & ge k^2
end{align}
How do I prove that $k^2=b^2+(a-c)^2$ ?
trigonometry
edited 7 hours ago
Ng Chung Tak
13.5k31234
asked 8 hours ago
ss1729
1,613722
1
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $
left|
asin^2theta+bsinthetacostheta+ccos^2theta-dfrac{(a+c)}{2}
right|=dfrac{k}{2}$, then find $k^2.$
My Attempt
begin{align}
pm k &= 2asin^2 theta+2bsin theta cos theta+2ccos^2 theta-(a+c) \
&= a[1-cos 2theta]+bsin 2theta+c[1+cos 2theta]-(a+c) \
&= aleft[ 1-frac{1-tan^2 theta}{1+tan^2 theta} right]+
bfrac{2tan theta}{1+tan^2 theta}+
cleft[ 1+frac{1-tan^2 theta}{1+tan^2 theta} right]-(a+c)\
pm k[1+tan^2 theta] &=
2atan^2 theta+2btan theta+2c-(a+c)[1+tan^2 theta] \
0 &= tan^2 theta [a-cmp k]+2btan theta+(c-amp k) \
Delta &= 4b^2-4(a-cmp k)(c-amp k) \
&= 4b^2-4[mp k+(a-c)][mp k-(a-c)] \
&= 4b^2-4[k^2-(a-c)^2] \
& ge 0 \
b^2+(a-c)^2 & ge k^2
end{align}
How do I prove that $k^2=b^2+(a-c)^2$ ?
trigonometry
edited 7 hours ago
Ng Chung Tak
13.5k31234
asked 8 hours ago
ss1729
1,613722
If $
left|
asin^2theta+bsinthetacostheta+ccos^2theta-dfrac{(a+c)}{2}
right|=dfrac{k}{2}$, then find $k^2.$
My Attempt
begin{align}
pm k &= 2asin^2 theta+2bsin theta cos theta+2ccos^2 theta-(a+c) \
&= a[1-cos 2theta]+bsin 2theta+c[1+cos 2theta]-(a+c) \
&= aleft[ 1-frac{1-tan^2 theta}{1+tan^2 theta} right]+
bfrac{2tan theta}{1+tan^2 theta}+
cleft[ 1+frac{1-tan^2 theta}{1+tan^2 theta} right]-(a+c)\
pm k[1+tan^2 theta] &=
2atan^2 theta+2btan theta+2c-(a+c)[1+tan^2 theta] \
0 &= tan^2 theta [a-cmp k]+2btan theta+(c-amp k) \
Delta &= 4b^2-4(a-cmp k)(c-amp k) \
&= 4b^2-4[mp k+(a-c)][mp k-(a-c)] \
&= 4b^2-4[k^2-(a-c)^2] \
& ge 0 \
b^2+(a-c)^2 & ge k^2
end{align}
How do I prove that $k^2=b^2+(a-c)^2$ ?
trigonometry
trigonometry
edited 7 hours ago
Ng Chung Tak
13.5k31234
asked 8 hours ago
ss1729
1,613722
edited 7 hours ago
Ng Chung Tak
13.5k31234
asked 8 hours ago
ss1729
1,613722
edited 7 hours ago
Ng Chung Tak
13.5k31234
edited 7 hours ago
Ng Chung Tak
13.5k31234
edited 7 hours ago
Ng Chung Tak
13.5k31234
13.5k31234
asked 8 hours ago
ss1729
1,613722
asked 8 hours ago
ss1729
1,613722
asked 8 hours ago
ss1729
1,613722
1,613722
1
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago
add a comment |
1
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago
1
1
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago
add a comment |
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1
$$2asin^2t+2bsin tcos t+2ccos^2t-(a+c)=cos2t(c-a)+bsin2tlesqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$
– lab bhattacharjee
8 hours ago
You can not prove $k^2=b^2+(a-c)^2$, because for $theta = 0$, $k^2=(c-a)^2$.
– farruhota
5 hours ago